L'Hopitals Rule

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April 13th 2008, 06:13 PM
dangerous_dave

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)

Please help :D
April 13th 2008, 06:17 PM
Mathstud28

Ok

Quote:

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)

Please help :D

L'hopital's is the EASIEST way to do this....i will do the hardest one for you $\lim_{x \to {\infty}}\frac{x}{ln\bigg(1+2e^{x}\bigg)}\Rightarr ow{\lim_{x \to {\infty}}\frac{1}{\frac{2e^{x}}{1+2e^{x}}}}\Righta rrow{1}$
April 13th 2008, 06:27 PM
dangerous_dave

Could you please explain how L'Hopitals works? I don't understand at all...
April 13th 2008, 06:44 PM
angel.white

Quote:

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 $\pm \frac \infty \infty$ or $\frac 00$ then the limit will be equal to the derivative of the numerator over the derivative of the denominator.

For example $\lim_{x\to 0} \frac {sin~x}{x}$ takes the form of 0/0 so we apply L'Hospital's rule:

$\lim_{x\to 0} \frac {sin~x}{x} ~~~~=~~~~ \lim_{x\to 0} \frac {\frac d{dx} sin~x}{\frac {dx}{dx}} ~~~~=~~~~ \lim_{x\to 0} \frac {cos~x}{1}~~~~=~~~~cos~0~~~~=~~~~1$
April 13th 2008, 07:01 PM
dangerous_dave

Wow thats a whole lot easier than other sites like to make out. Thanks heaps :D
April 13th 2008, 07:28 PM
ThePerfectHacker

Does "Dangerous Dave" come from the 1990 computer game?

Quote:

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 $\ln x$ at $x=1$ .

Quote:

c) lim x->0 (cot2x)(sin6x)

$\cot 2x \sin 6x = \cos 2x \left( \frac{\sin 6x}{\sin 2x} \right)$ .
Note, $\cos 2x \to 1$ .
While, $\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{2x}{\sin 2x}\cdot 3\to 1\cdot 1\cdot 3 = 3$