text: 1. ## L'Hopitals Rule
  
  I have these 3 questions, and I think I need to use L'Hopitals rule. But I Have no idea how to go about that. Anything I've read about it confuses me.
  
  a)
  lim x->1 (ln x)/(x-1)
  
  b) lim x-> infinity x/ln(1+2e^x)
  
  c) lim x->0 (cot2x)(sin6x)
  
  2. ## Ok
  
  Originally Posted by dangerous_dave
  I have these 3 questions, and I think I need to use L'Hopitals rule. But I Have no idea how to go about that. Anything I've read about it confuses me.
  
  a)
  lim x->1 (ln x)/(x-1)
  
  b) lim x-> infinity x/ln(1+2e^x)
  
  c) lim x->0 (cot2x)(sin6x)
  
  L'hopital's is the EASIEST way to do this....i will do the hardest one for you $\displaystyle \lim_{x \to {\infty}}\frac{x}{ln\bigg(1+2e^{x}\bigg)}\Rightarr ow{\lim_{x \to {\infty}}\frac{1}{\frac{2e^{x}}{1+2e^{x}}}}\Righta rrow{1}$
  
  3. Could you please explain how L'Hopitals works? I don't understand at all...
  
  4. Originally Posted by dangerous_dave
  Could you please explain how L'Hopitals works? I don't understand at all...
  If you have a limit which takes the form of $\displaystyle \pm \frac \infty \infty$ or $\displaystyle \frac 00$ then the limit will be equal to the derivative of the numerator over the derivative of the denominator.
  
  For example $\displaystyle \lim_{x\to 0} \frac {sin~x}{x}$ takes the form of 0/0 so we apply L'Hospital's rule:
  
  $\displaystyle \lim_{x\to 0} \frac {sin~x}{x} ~~~~=~~~~ \lim_{x\to 0} \frac {\frac d{dx} sin~x}{\frac {dx}{dx}} ~~~~=~~~~ \lim_{x\to 0} \frac {cos~x}{1}~~~~=~~~~cos~0~~~~=~~~~1$
  
  5. Wow thats a whole lot easier than other sites like to make out. Thanks heaps
  
  6. Does "Dangerous Dave" come from the 1990 computer game?
  
  Originally Posted by dangerous_dave
  a)
  lim x->1 (ln x)/(x-1)
  This is nothing to do with L'Hopital rule. This is the definition of derivative of $\displaystyle \ln x$ at $\displaystyle x=1$.
  
  c) lim x->0 (cot2x)(sin6x)
  $\displaystyle \cot 2x \sin 6x = \cos 2x \left( \frac{\sin 6x}{\sin 2x} \right)$.
  Note, $\displaystyle \cos 2x \to 1$.
  While, $\displaystyle \frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{2x}{\sin 2x}\cdot 3\to 1\cdot 1\cdot 3 = 3$
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