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