text: Informative line
  
  ### L Hospital Rule
  
  Learn L’Hospital’s rule function and evaluating L’hospital’s Rule by exponential, logarithmic or trigonometric function.
  
  # L' Hospital Rule(0/0 form)
  
  ## Indeterminate form $$\Big(\dfrac {0}{0}\Big)$$
  
  • A limit of the form $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$ where $$f$$ and $$g$$are any two function of $$x$$is said to be of $$\dfrac {0}{0}$$form if $$\lim\limits_{x\to a}\,f(x)=0$$ and $$\lim\limits_{x\to a}\,g(x)=0$$
  • When $$f$$ and $$g$$are polynomial or algebraic functions the limits can be evaluated by elementary methods but sometimes $$f, \,g$$ contains exponential, logarithmic or trigonometric function.
  • For these limits we use L' Hospital Rule which says, that
  • If $$f$$ and $$g$$are differentiable and $$g'(x)\neq0$$ on an open interval  $$I$$ , that contains $$'a'$$. Suppose that
  
  $$\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=0$$ then
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}$$
  
  If the limits on the right side exist.
  
  • So to evaluate $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$suppose we observe that it is $$\dfrac {0}{0}$$ form then we use
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}$$
  
  If $$\dfrac {0}{0}$$ form is removed, the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {1-sin\,\theta}{1+cos2\,\theta}$$ .
  
  A $$\dfrac {1}{4}$$
  
  B $$\dfrac {1}{2}$$
  
  C $$\dfrac {3}{2}$$
  
  D $$\dfrac {3}{4}$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=0$$
  
  In this case the limit is
  
  $$\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)$$ which is $$\Big(\dfrac {0}{0}\Big)$$form
  
  $$\therefore \;\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)= \lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {\dfrac {d}{d\theta}(1-sin\,\theta)} {\dfrac {d}{d\theta}(1+cos2\,\theta)}$$
  
  $$=\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {-cos\,\theta}{-2sin2\,\theta}= \lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {cos\,\theta} {2\,sin2\,\theta}$$
  
  Put $$\theta\to\dfrac {\pi}{2}$$ we get $$\dfrac {0}{0}$$ form again
  
  $$\therefore \,\,\lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {cos\,\theta}{2\,sin2\,\theta}= \lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {-sin\,\theta} {4\,cos2\,\theta}= \dfrac {-sin\dfrac {\pi}{2}} {4\,cos\,\pi}=\dfrac {1}{4}$$
  
  $$\therefore \;\;\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)=\dfrac {1}{4}$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {1-sin\,\theta}{1+cos2\,\theta}$$ .
  
  A
  
  $$\dfrac {1}{4}$$
  
  .
  
  B
  
  $$\dfrac {1}{2}$$
  
  C
  
  $$\dfrac {3}{2}$$
  
  D
  
  $$\dfrac {3}{4}$$
  
  Option A is Correct
  
  # Indeterminate Form  (0/0 form)
  
  • A limit of the form $$\lim\limits_{x\to a}\;\dfrac {f(x)}{g(x)}$$ where $$f$$ and  $$g$$are any two functions of $$x$$, is said to be of $$\left (\dfrac {0}{0}\right)$$form if  $$\lim\limits_{x\to a}\;f(x)=0$$ and $$\lim\limits_{x\to a}\;g(x)=0$$.
  • When $$f$$ and  $$g$$are polynomial or algebraic functions, the limit can be evaluated by elementary method, but sometimes $$f$$$$g$$ contain exponential, logarithmic and trigonometric functions.
  • For these limits, we use L' Hospital rule.
  
  #### In which one of the following L' Hospital rule can be applied?
  
  A $$\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  B $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  C $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}$$
  
  D $$\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}$$
  
  ×
  
  In option A
  
  $$\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  $$\Rightarrow\;\dfrac {1-sin\,\dfrac {\pi}{2}}{1+cos\,2×\dfrac {\pi}{2}}$$
  
  $$\Rightarrow\;\dfrac {1-1}{1-1}$$
  
  $$\Rightarrow\;\dfrac {0}{0}$$ from
  
  Hence, L' Hospital rule can be applied.
  
  Thus option (A) is correct.
  
  In option B
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  $$=\dfrac {1-sin\,0}{1+cos\,0}$$
  
  $$=\dfrac {1}{2}$$
  
  Thus here L' Hospital rule can not be applied as no $$\dfrac{0}{0}$$ form is obtained.
  
  Hence, option (B) is incorrect.
  
  In option C
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}$$
  
  $$=\dfrac {1-sin\,0}{1+cos\,0}$$
  
  $$=\dfrac {1}{2}$$
  
  Thus here L' Hospital rule can not be applied as no $$\dfrac{0}{0}$$ form is obtained.
  
  Hence, option (C) is incorrect.
  
  In option D
  
  $$\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}$$
  
  $$=\dfrac {1-sin\,\pi}{1+sin\,\pi}$$
  
  $$=\dfrac {1}{1}=1$$
  
  Thus L' Hospital rule can not be applied as no $$\dfrac{0}{0}$$ form is obtained.
  
  Hence option (D) is incorrect.
  
  ### In which one of the following L' Hospital rule can be applied?
  
  A
  
  $$\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  .
  
  B
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  C
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}$$
  
  D
  
  $$\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}$$
  
  Option A is Correct
  
  # L' Hospital Rule for Logarithmic Function (infinite by infinite form)
  
  ## Indeterminate form $$\Big(\dfrac {\infty}{\infty}\Big)$$
  
  • A limit of the form $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$ where $$f$$ and $$g$$are any two function of $$x$$is said to be of $$\Big(\dfrac {\infty}{\infty}\Big)$$form if $$\lim\limits_{x\to a}\,f(x)=\infty$$ and $$\lim\limits_{x\to a}\,g(x)=\infty$$
  • When $$f$$ and $$g$$are polynomial or algebraic functions the limits can be evaluated by elementary methods but sometimes $$f, \,g$$ contains exponential, logarithmic or trigonometric function.
  • For these limits we use L' Hospital Rule which says, that
  • If $$f$$ and $$g$$are differentiable and $$g'(x)\neq \infty$$ on an open interval $$I$$, that contains $$'a'$$. Suppose that
  
  $$\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=\infty$$ then
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}$$
  
  if the limits on the right side exist.
  
  • So to evaluate $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$suppose we observe that it is $$\Big(\dfrac {\infty}{\infty}\Big)$$ form then we use
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\to$$If $$\Big(\dfrac {\infty}{\infty}\Big)$$ form is removed the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)$$ .
  
  A $$\dfrac {1}{2}$$
  
  B $$0$$
  
  C $$\ell n\,2$$
  
  D $$1$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=\infty$$
  
  In this case the limit is
  
  $$\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)$$ which is $$\Big(\dfrac {\infty}{\infty}\Big)$$form
  
  $$\therefore \; \lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big) = \lim\limits_{x \to \infty}\;\dfrac {\dfrac {d}{dx}(\ell n\,x)} {\dfrac {d}{dx}(\sqrt x)}$$
  
  $$=\lim\limits_{x\to \infty}\; \dfrac {\dfrac {1}{x}} {\dfrac {1}{2}x^{-1/2}}$$
  
  $$=\lim\limits_{x\to\infty}\;\dfrac {2\sqrt x}{x}= \lim\limits_{x\to\infty}\;\dfrac {2}{\sqrt x}=0$$
  
  $$\therefore \; \lim\limits_{x\to\infty}\;\dfrac {\ell n\,x}{\sqrt x}=0$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)$$ .
  
  A
  
  $$\dfrac {1}{2}$$
  
  .
  
  B
  
  $$0$$
  
  C
  
  $$\ell n\,2$$
  
  D
  
  $$1$$
  
  Option B is Correct
  
  # Indeterminate Product Form
  
  ## Indeterminate Form$$(0\times \infty)$$
  
  • $$\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)$$ is said to be of the form $$0×\infty$$. If $$\lim\limits_{x\to a}\,f(x)=0$$ and $$\lim\limits_{x\to a}\,g(x)=\infty$$ (or $$-\infty$$).
  • If the above (limit cannot be evaluated by elementary methods we find expression $$\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)$$ as $$=\lim\limits_{x \to a}\dfrac {f(x)}{1/g\,(x)}$$ or $$\lim\limits_{x \to a}\dfrac {g(x)}{1/f\,(x)}$$ and then use L' Hospital rule.
  
  #### Evaluate $$\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)$$ .
  
  A $$\dfrac {2}{\pi}$$
  
  B $$\dfrac {-2}{\pi}$$
  
  C $$\dfrac {\pi}{2}$$
  
  D $$\dfrac {-\pi}{2}$$
  
  ×
  
  $$\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)=\lim\limits_{x \to a}\dfrac {f(x)}{1/g\,(x)}$$ or $$\lim\limits_{x \to a}\dfrac {g(x)}{1/f\,(x)}$$
  
  and then apply L' Hospital rule (for $$0×\infty$$ form)
  
  In this case the limit is
  
  $$\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)= \lim\limits_{x \to 1}\;\dfrac {(\ell n\,x)}{\left (cot\,\dfrac {\pi\,x}{2}\right)}$$
  
  which is of $$\left (\dfrac {0}{0}\right)$$form.
  
  $$\therefore \lim\limits_{x \to 1}\;\dfrac {(\ell n\,x)}{\left (cot\,\dfrac {\pi\,x}{2}\right)}$$
  
  $$= \lim\limits_{x \to 1}\;\dfrac {\dfrac {d}{dx}(\ell n\,x)}{\dfrac {d}{dx}\left (cot\,\dfrac {\pi\,x}{2}\right)}$$
  
  $$= \lim\limits_{x \to 1}\;\dfrac {\dfrac {1}{x}}{-cosec^2\,\dfrac {\pi\,x}{2}×\dfrac {\pi}{2}}$$
  
  $$=\dfrac {-2}{\pi}$$
  
  $$\therefore \lim\limits_{x\to 1}\,(\ell n\,x)\Big(tan\,\dfrac{\pi\,x}{2}\Big)=\dfrac {-2}{\pi}$$
  
  ### Evaluate $$\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)$$ .
  
  A
  
  $$\dfrac {2}{\pi}$$
  
  .
  
  B
  
  $$\dfrac {-2}{\pi}$$
  
  C
  
  $$\dfrac {\pi}{2}$$
  
  D
  
  $$\dfrac {-\pi}{2}$$
  
  Option B is Correct
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 0}\;\dfrac {x\,2^x}{2^x-1}$$ .
  
  A $$\dfrac {1}{2\, \ell n\,2}$$
  
  B $$2\,\ell n\,2$$
  
  C $$\ell n\,2$$
  
  D $$\dfrac {1}{\ell n\,2}$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=0$$
  
  In this case the limit is
  
  $$\lim\limits_{x \to 0}\;\left (\dfrac {x\,2^x}{2^x-1}\right)$$ which is $$\Big(\dfrac {0}{0}\Big)$$form
  
  $$\therefore \; \lim\limits_{x \to 0}\;\left (\dfrac {x\,2^x}{2^x-1}\right)= \lim\limits_{x \to 0}\; \dfrac {\dfrac {d}{dx}(x\,2^x)}{\dfrac {d}{dx}(2^x-1)}$$
  
  $$= \lim\limits_{x \to 0}\; \dfrac {x×2^x\,\ell n\,2+2^x}{2^x\,\ell n\,2}$$ (Now put $$x = 0$$ )
  
  $$=\dfrac {0+1}{\ell n\,2}=\dfrac {1}{\ell n\,2}$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 0}\;\dfrac {x\,2^x}{2^x-1}$$ .
  
  A
  
  $$\dfrac {1}{2\, \ell n\,2}$$
  
  .
  
  B
  
  $$2\,\ell n\,2$$
  
  C
  
  $$\ell n\,2$$
  
  D
  
  $$\dfrac {1}{\ell n\,2}$$
  
  Option D is Correct
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x} \right)$$ .
  
  A $$-\pi$$
  
  B $$-\dfrac {1}{\pi^2}$$
  
  C $$\dfrac {1}{\pi^2}$$
  
  D $$\pi$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=0$$
  
  In this case the limit is
  
  $$\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right)$$ which is $$\Big(\dfrac {0}{0}\Big)$$form
  
  $$\therefore \; \lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right) = \lim\limits_{x \to 1}\; \dfrac {\dfrac {d}{dx}(1-x+\ell n\,x)}{\dfrac {d}{dx}(1+cos\,\pi\,x)}$$
  
  $$=\lim\limits_{x \to 1}\;\left (\dfrac {-1+1/x}{-\pi\,\sin\,\pi\,x}\right)$$
  
  Put $$x=1$$, we get  $$\Big(\dfrac {0}{0}\Big)$$form again
  
  $$\therefore \;\lim\limits_{x \to 1}\;\left (\dfrac {-1+1/x}{-\pi\,\sin\,\pi\,x}\right)$$
  
  $$=\lim\limits_{x \to 1}\;\dfrac {\dfrac {d}{dx}(-1+1/x)} {\dfrac {d}{dx}(-\pi\,\sin\,\pi\,x)}$$
  
  $$=\lim\limits_{x \to 1}\; \dfrac {\dfrac {-1}{x^2}}{-\pi^2\,cos\,\pi\,x}=\dfrac {-1}{\pi^2}$$
  
  $$\therefore, \;\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right)=\dfrac {-1}{\pi^2}$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x} \right)$$ .
  
  A
  
  $$-\pi$$
  
  .
  
  B
  
  $$-\dfrac {1}{\pi^2}$$
  
  C
  
  $$\dfrac {1}{\pi^2}$$
  
  D
  
  $$\pi$$
  
  Option B is Correct
  
  # Indeterminate Power Forms
  
  The Limits of the form $$\lim\limits_{x\to a}\;\Big(f(x)\Big)^{g(x)}$$ can be of any one of the following indeterminate form
  
  1. $$1^{\infty}\to$$ when $$\lim\limits_{x\to a}\; f(x)=1$$ and $$\lim\limits_{x\to a}\; g(x)=\infty$$
  2. $$0^0\to$$ when $$\lim\limits_{x\to a}\; f(x)= \lim\limits_{x\to a}\; g(x)=0$$
  3. $$\infty^0\to$$form when $$\lim\limits_{x\to a}\; f(x)=\infty$$ & $$\lim\limits_{x\to a}\; g(x)=0$$
  
  Each of the these limits can be evaluated by assuming the limits to be y and the taking log on both sides and applying L' Hospital rule.
  
  #### Evaluate $$\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}$$ using L' Hospital Rule.
  
  A $$e^{(-2/\pi)}$$
  
  B $$e^{(2/\pi)}$$
  
  C $$e^{(\pi/2)}$$
  
  D $$e^{(-\pi/2)}$$
  
  ×
  
  For evaluating $$\lim\limits_{x \to a}\,\big(f(x)\big)^{g(x)}$$ we first take log and then use L' Hospital Rule.
  
  Let
  
  $$y=\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}$$ which is of $$1^{\infty}$$ form.
  
  Take log on both sides $$\Rightarrow$$$$log\,y=\lim\limits_{x\to 1}\; {\Big(tan\dfrac {\pi\, x}{2}\Big)}\;\big(\ell n\,(2-x)\big)$$
  
  $$\Rightarrow$$ $$log\,y=\lim\limits_{x\to 1}\; \dfrac {\ell n\,(2-x)}{{\Big(cot\dfrac {\pi\, x}{2}\Big)}\;}$$
  
  Apply L' Hospital Rule
  
  $$log\,y=\lim\limits_{x\to 1}\; \dfrac {\dfrac {d}{dx}\ell n\,(2-x)} {\dfrac {d}{dx}{\Big(cot\dfrac {\pi\, x}{2}\Big)}\;}$$
  
  $$=\lim\limits_{x\to 1}\; \dfrac {\dfrac {1}{2-x}×(-1)} {-cosec^2\,\dfrac {\pi\,x}{2}×\dfrac {\pi}{2}\;}$$
  
  $$=\dfrac {-1}{-1×\dfrac {\pi}{2}}=\dfrac{2} {\pi}$$
  
  $$\Rightarrow\;y=e^{(2/\pi)}$$
  
  ### Evaluate $$\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}$$ using L' Hospital Rule.
  
  A
  
  $$e^{(-2/\pi)}$$
  
  .
  
  B
  
  $$e^{(2/\pi)}$$
  
  C
  
  $$e^{(\pi/2)}$$
  
  D
  
  $$e^{(-\pi/2)}$$
  
  Option B is Correct
explanation: The provided text is a detailed explanation of L'Hospital's Rule and its application to various indeterminate forms, such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), and indeterminate power forms. The text includes several examples and solutions to problems involving these indeterminate forms.
  
  The text contains multiple questions and their corresponding answers. Each question is followed by a detailed solution, and the correct answer is highlighted at the end.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: Informative line
  
  ### L Hospital Rule
  
  Learn L’Hospital’s rule function and evaluating L’hospital’s Rule by exponential, logarithmic or trigonometric function.
  
  # L' Hospital Rule(0/0 form)
  
  ## Indeterminate form $$\Big(\dfrac {0}{0}\Big)$$
  
  • A limit of the form $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$ where $$f$$ and $$g$$are any two function of $$x$$is said to be of $$\dfrac {0}{0}$$form if $$\lim\limits_{x\to a}\,f(x)=0$$ and $$\lim\limits_{x\to a}\,g(x)=0$$
  • When $$f$$ and $$g$$are polynomial or algebraic functions the limits can be evaluated by elementary methods but sometimes $$f, \,g$$ contains exponential, logarithmic or trigonometric function.
  • For these limits we use L' Hospital Rule which says, that
  • If $$f$$ and $$g$$are differentiable and $$g'(x)\neq0$$ on an open interval  $$I$$ , that contains $$'a'$$. Suppose that
  
  $$\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=0$$ then
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}$$
  
  If the limits on the right side exist.
  
  • So to evaluate $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$suppose we observe that it is $$\dfrac {0}{0}$$ form then we use
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}$$
  
  If $$\dfrac {0}{0}$$ form is removed, the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {1-sin\,\theta}{1+cos2\,\theta}$$ .
  
  A $$\dfrac {1}{4}$$
  
  B $$\dfrac {1}{2}$$
  
  C $$\dfrac {3}{2}$$
  
  D $$\dfrac {3}{4}$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=0$$
  
  In this case the limit is
  
  $$\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)$$ which is $$\Big(\dfrac {0}{0}\Big)$$form
  
  $$\therefore \;\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)= \lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {\dfrac {d}{d\theta}(1-sin\,\theta)} {\dfrac {d}{d\theta}(1+cos2\,\theta)}$$
  
  $$=\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {-cos\,\theta}{-2sin2\,\theta}= \lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {cos\,\theta} {2\,sin2\,\theta}$$
  
  Put $$\theta\to\dfrac {\pi}{2}$$ we get $$\dfrac {0}{0}$$ form again
  
  $$\therefore \,\,\lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {cos\,\theta}{2\,sin2\,\theta}= \lim\limits_{\theta\to\dfrac {\pi}{2}}\; \dfrac {-sin\,\theta} {4\,cos2\,\theta}= \dfrac {-sin\dfrac {\pi}{2}} {4\,cos\,\pi}=\dfrac {1}{4}$$
  
  $$\therefore \;\;\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\left (\dfrac {1-sin\,\theta}{1+cos2\,\theta}\right)=\dfrac {1}{4}$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{\theta\to\dfrac {\pi}{2}}\;\dfrac {1-sin\,\theta}{1+cos2\,\theta}$$ .
  
  A
  
  $$\dfrac {1}{4}$$
  
  .
  
  B
  
  $$\dfrac {1}{2}$$
  
  C
  
  $$\dfrac {3}{2}$$
  
  D
  
  $$\dfrac {3}{4}$$
  
  Option A is Correct
  
  # Indeterminate Form  (0/0 form)
  
  • A limit of the form $$\lim\limits_{x\to a}\;\dfrac {f(x)}{g(x)}$$ where $$f$$ and  $$g$$are any two functions of $$x$$, is said to be of $$\left (\dfrac {0}{0}\right)$$form if  $$\lim\limits_{x\to a}\;f(x)=0$$ and $$\lim\limits_{x\to a}\;g(x)=0$$.
  • When $$f$$ and  $$g$$are polynomial or algebraic functions, the limit can be evaluated by elementary method, but sometimes $$f$$$$g$$ contain exponential, logarithmic and trigonometric functions.
  • For these limits, we use L' Hospital rule.
  
  #### In which one of the following L' Hospital rule can be applied?
  
  A $$\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  B $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  C $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}$$
  
  D $$\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}$$
  
  ×
  
  In option A
  
  $$\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  $$\Rightarrow\;\dfrac {1-sin\,\dfrac {\pi}{2}}{1+cos\,2×\dfrac {\pi}{2}}$$
  
  $$\Rightarrow\;\dfrac {1-1}{1-1}$$
  
  $$\Rightarrow\;\dfrac {0}{0}$$ from
  
  Hence, L' Hospital rule can be applied.
  
  Thus option (A) is correct.
  
  In option B
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  $$=\dfrac {1-sin\,0}{1+cos\,0}$$
  
  $$=\dfrac {1}{2}$$
  
  Thus here L' Hospital rule can not be applied as no $$\dfrac{0}{0}$$ form is obtained.
  
  Hence, option (B) is incorrect.
  
  In option C
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}$$
  
  $$=\dfrac {1-sin\,0}{1+cos\,0}$$
  
  $$=\dfrac {1}{2}$$
  
  Thus here L' Hospital rule can not be applied as no $$\dfrac{0}{0}$$ form is obtained.
  
  Hence, option (C) is incorrect.
  
  In option D
  
  $$\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}$$
  
  $$=\dfrac {1-sin\,\pi}{1+sin\,\pi}$$
  
  $$=\dfrac {1}{1}=1$$
  
  Thus L' Hospital rule can not be applied as no $$\dfrac{0}{0}$$ form is obtained.
  
  Hence option (D) is incorrect.
  
  ### In which one of the following L' Hospital rule can be applied?
  
  A
  
  $$\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  .
  
  B
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}$$
  
  C
  
  $$\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}$$
  
  D
  
  $$\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}$$
  
  Option A is Correct
  
  # L' Hospital Rule for Logarithmic Function (infinite by infinite form)
  
  ## Indeterminate form $$\Big(\dfrac {\infty}{\infty}\Big)$$
  
  • A limit of the form $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$ where $$f$$ and $$g$$are any two function of $$x$$is said to be of $$\Big(\dfrac {\infty}{\infty}\Big)$$form if $$\lim\limits_{x\to a}\,f(x)=\infty$$ and $$\lim\limits_{x\to a}\,g(x)=\infty$$
  • When $$f$$ and $$g$$are polynomial or algebraic functions the limits can be evaluated by elementary methods but sometimes $$f, \,g$$ contains exponential, logarithmic or trigonometric function.
  • For these limits we use L' Hospital Rule which says, that
  • If $$f$$ and $$g$$are differentiable and $$g'(x)\neq \infty$$ on an open interval $$I$$, that contains $$'a'$$. Suppose that
  
  $$\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=\infty$$ then
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}$$
  
  if the limits on the right side exist.
  
  • So to evaluate $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)}$$suppose we observe that it is $$\Big(\dfrac {\infty}{\infty}\Big)$$ form then we use
  
  $$\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\to$$If $$\Big(\dfrac {\infty}{\infty}\Big)$$ form is removed the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)$$ .
  
  A $$\dfrac {1}{2}$$
  
  B $$0$$
  
  C $$\ell n\,2$$
  
  D $$1$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=\infty$$
  
  In this case the limit is
  
  $$\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)$$ which is $$\Big(\dfrac {\infty}{\infty}\Big)$$form
  
  $$\therefore \; \lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big) = \lim\limits_{x \to \infty}\;\dfrac {\dfrac {d}{dx}(\ell n\,x)} {\dfrac {d}{dx}(\sqrt x)}$$
  
  $$=\lim\limits_{x\to \infty}\; \dfrac {\dfrac {1}{x}} {\dfrac {1}{2}x^{-1/2}}$$
  
  $$=\lim\limits_{x\to\infty}\;\dfrac {2\sqrt x}{x}= \lim\limits_{x\to\infty}\;\dfrac {2}{\sqrt x}=0$$
  
  $$\therefore \; \lim\limits_{x\to\infty}\;\dfrac {\ell n\,x}{\sqrt x}=0$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{x\to \infty}\;\Big(\dfrac {\ell n\,x}{\sqrt x}\Big)$$ .
  
  A
  
  $$\dfrac {1}{2}$$
  
  .
  
  B
  
  $$0$$
  
  C
  
  $$\ell n\,2$$
  
  D
  
  $$1$$
  
  Option B is Correct
  
  # Indeterminate Product Form
  
  ## Indeterminate Form$$(0\times \infty)$$
  
  • $$\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)$$ is said to be of the form $$0×\infty$$. If $$\lim\limits_{x\to a}\,f(x)=0$$ and $$\lim\limits_{x\to a}\,g(x)=\infty$$ (or $$-\infty$$).
  • If the above (limit cannot be evaluated by elementary methods we find expression $$\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)$$ as $$=\lim\limits_{x \to a}\dfrac {f(x)}{1/g\,(x)}$$ or $$\lim\limits_{x \to a}\dfrac {g(x)}{1/f\,(x)}$$ and then use L' Hospital rule.
  
  #### Evaluate $$\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)$$ .
  
  A $$\dfrac {2}{\pi}$$
  
  B $$\dfrac {-2}{\pi}$$
  
  C $$\dfrac {\pi}{2}$$
  
  D $$\dfrac {-\pi}{2}$$
  
  ×
  
  $$\lim\limits_{x\to a}\Big(f(x)\,g(x)\Big)=\lim\limits_{x \to a}\dfrac {f(x)}{1/g\,(x)}$$ or $$\lim\limits_{x \to a}\dfrac {g(x)}{1/f\,(x)}$$
  
  and then apply L' Hospital rule (for $$0×\infty$$ form)
  
  In this case the limit is
  
  $$\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)= \lim\limits_{x \to 1}\;\dfrac {(\ell n\,x)}{\left (cot\,\dfrac {\pi\,x}{2}\right)}$$
  
  which is of $$\left (\dfrac {0}{0}\right)$$form.
  
  $$\therefore \lim\limits_{x \to 1}\;\dfrac {(\ell n\,x)}{\left (cot\,\dfrac {\pi\,x}{2}\right)}$$
  
  $$= \lim\limits_{x \to 1}\;\dfrac {\dfrac {d}{dx}(\ell n\,x)}{\dfrac {d}{dx}\left (cot\,\dfrac {\pi\,x}{2}\right)}$$
  
  $$= \lim\limits_{x \to 1}\;\dfrac {\dfrac {1}{x}}{-cosec^2\,\dfrac {\pi\,x}{2}×\dfrac {\pi}{2}}$$
  
  $$=\dfrac {-2}{\pi}$$
  
  $$\therefore \lim\limits_{x\to 1}\,(\ell n\,x)\Big(tan\,\dfrac{\pi\,x}{2}\Big)=\dfrac {-2}{\pi}$$
  
  ### Evaluate $$\lim\limits_{x \to 1}\;(\ell n\,x)\left (tan\,\dfrac {\pi\,x}{2}\right)$$ .
  
  A
  
  $$\dfrac {2}{\pi}$$
  
  .
  
  B
  
  $$\dfrac {-2}{\pi}$$
  
  C
  
  $$\dfrac {\pi}{2}$$
  
  D
  
  $$\dfrac {-\pi}{2}$$
  
  Option B is Correct
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 0}\;\dfrac {x\,2^x}{2^x-1}$$ .
  
  A $$\dfrac {1}{2\, \ell n\,2}$$
  
  B $$2\,\ell n\,2$$
  
  C $$\ell n\,2$$
  
  D $$\dfrac {1}{\ell n\,2}$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=0$$
  
  In this case the limit is
  
  $$\lim\limits_{x \to 0}\;\left (\dfrac {x\,2^x}{2^x-1}\right)$$ which is $$\Big(\dfrac {0}{0}\Big)$$form
  
  $$\therefore \; \lim\limits_{x \to 0}\;\left (\dfrac {x\,2^x}{2^x-1}\right)= \lim\limits_{x \to 0}\; \dfrac {\dfrac {d}{dx}(x\,2^x)}{\dfrac {d}{dx}(2^x-1)}$$
  
  $$= \lim\limits_{x \to 0}\; \dfrac {x×2^x\,\ell n\,2+2^x}{2^x\,\ell n\,2}$$ (Now put $$x = 0$$ )
  
  $$=\dfrac {0+1}{\ell n\,2}=\dfrac {1}{\ell n\,2}$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 0}\;\dfrac {x\,2^x}{2^x-1}$$ .
  
  A
  
  $$\dfrac {1}{2\, \ell n\,2}$$
  
  .
  
  B
  
  $$2\,\ell n\,2$$
  
  C
  
  $$\ell n\,2$$
  
  D
  
  $$\dfrac {1}{\ell n\,2}$$
  
  Option D is Correct
  
  #### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x} \right)$$ .
  
  A $$-\pi$$
  
  B $$-\dfrac {1}{\pi^2}$$
  
  C $$\dfrac {1}{\pi^2}$$
  
  D $$\pi$$
  
  ×
  
  According to L' Hospital Rule
  
  $$\lim\limits_{x\to a}\dfrac {f(x)}{g(x)} =\lim\limits_{x\to a}\dfrac {f'(x)}{g'(x)}$$
  
  if $$f(a)=g(a)=0$$
  
  In this case the limit is
  
  $$\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right)$$ which is $$\Big(\dfrac {0}{0}\Big)$$form
  
  $$\therefore \; \lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right) = \lim\limits_{x \to 1}\; \dfrac {\dfrac {d}{dx}(1-x+\ell n\,x)}{\dfrac {d}{dx}(1+cos\,\pi\,x)}$$
  
  $$=\lim\limits_{x \to 1}\;\left (\dfrac {-1+1/x}{-\pi\,\sin\,\pi\,x}\right)$$
  
  Put $$x=1$$, we get  $$\Big(\dfrac {0}{0}\Big)$$form again
  
  $$\therefore \;\lim\limits_{x \to 1}\;\left (\dfrac {-1+1/x}{-\pi\,\sin\,\pi\,x}\right)$$
  
  $$=\lim\limits_{x \to 1}\;\dfrac {\dfrac {d}{dx}(-1+1/x)} {\dfrac {d}{dx}(-\pi\,\sin\,\pi\,x)}$$
  
  $$=\lim\limits_{x \to 1}\; \dfrac {\dfrac {-1}{x^2}}{-\pi^2\,cos\,\pi\,x}=\dfrac {-1}{\pi^2}$$
  
  $$\therefore, \;\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x}\right)=\dfrac {-1}{\pi^2}$$
  
  ### Use L' Hospital Rule to evaluate $$\lim\limits_{x \to 1}\;\left (\dfrac {1-x+\ell n\,x}{1+cos\,\pi\,x} \right)$$ .
  
  A
  
  $$-\pi$$
  
  .
  
  B
  
  $$-\dfrac {1}{\pi^2}$$
  
  C
  
  $$\dfrac {1}{\pi^2}$$
  
  D
  
  $$\pi$$
  
  Option B is Correct
  
  # Indeterminate Power Forms
  
  The Limits of the form $$\lim\limits_{x\to a}\;\Big(f(x)\Big)^{g(x)}$$ can be of any one of the following indeterminate form
  
  1. $$1^{\infty}\to$$ when $$\lim\limits_{x\to a}\; f(x)=1$$ and $$\lim\limits_{x\to a}\; g(x)=\infty$$
  2. $$0^0\to$$ when $$\lim\limits_{x\to a}\; f(x)= \lim\limits_{x\to a}\; g(x)=0$$
  3. $$\infty^0\to$$form when $$\lim\limits_{x\to a}\; f(x)=\infty$$ & $$\lim\limits_{x\to a}\; g(x)=0$$
  
  Each of the these limits can be evaluated by assuming the limits to be y and the taking log on both sides and applying L' Hospital rule.
  
  #### Evaluate $$\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}$$ using L' Hospital Rule.
  
  A $$e^{(-2/\pi)}$$
  
  B $$e^{(2/\pi)}$$
  
  C $$e^{(\pi/2)}$$
  
  D $$e^{(-\pi/2)}$$
  
  ×
  
  For evaluating $$\lim\limits_{x \to a}\,\big(f(x)\big)^{g(x)}$$ we first take log and then use L' Hospital Rule.
  
  Let
  
  $$y=\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}$$ which is of $$1^{\infty}$$ form.
  
  Take log on both sides $$\Rightarrow$$$$log\,y=\lim\limits_{x\to 1}\; {\Big(tan\dfrac {\pi\, x}{2}\Big)}\;\big(\ell n\,(2-x)\big)$$
  
  $$\Rightarrow$$ $$log\,y=\lim\limits_{x\to 1}\; \dfrac {\ell n\,(2-x)}{{\Big(cot\dfrac {\pi\, x}{2}\Big)}\;}$$
  
  Apply L' Hospital Rule
  
  $$log\,y=\lim\limits_{x\to 1}\; \dfrac {\dfrac {d}{dx}\ell n\,(2-x)} {\dfrac {d}{dx}{\Big(cot\dfrac {\pi\, x}{2}\Big)}\;}$$
  
  $$=\lim\limits_{x\to 1}\; \dfrac {\dfrac {1}{2-x}×(-1)} {-cosec^2\,\dfrac {\pi\,x}{2}×\dfrac {\pi}{2}\;}$$
  
  $$=\dfrac {-1}{-1×\dfrac {\pi}{2}}=\dfrac{2} {\pi}$$
  
  $$\Rightarrow\;y=e^{(2/\pi)}$$
  
  ### Evaluate $$\lim\limits_{x\to 1}\; (2-x)^{\Big(tan\dfrac {\pi\, x}{2}\Big)}$$ using L' Hospital Rule.
  
  A
  
  $$e^{(-2/\pi)}$$
  
  .
  
  B
  
  $$e^{(2/\pi)}$$
  
  C
  
  $$e^{(\pi/2)}$$
  
  D
  
  $$e^{(-\pi/2)}$$
  
  Option B is Correct
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a detailed explanation of L'Hospital's Rule and its application to various indeterminate forms, such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), and indeterminate power forms. The text includes several examples and solutions to problems involving these indeterminate forms.
  
  The text contains multiple questions and their corresponding answers. Each question is followed by a detailed solution, and the correct answer is highlighted at the end.
  
  Label: multiple-question-and-answer