# L'Hopital's rule for solving limit

mousesgr
1. lim [(1+x)^(1/x) - e ] / x
x ->0

2. lim [sin(2/x)+cos(1/x)]^x
x -> inf

help...

Last edited:

## Answers and Replies

Science Advisor
Homework Helper
Homework should be posted in the homework forum- and you should show us what you have tried to do. Have you considered L'Hopital's rule?

mousesgr
qs 1 is not 0/0 or inf/inf from
how do consider L'Hopital's rule?

Science Advisor
Homework Helper
Do you KNOW "L'Hopital's rule"? It is specifically for limit problems where you get things like these.

If you have $$lim_{x->a} \frac{f(x)}{g{x}}$$ where f(a)= 0 and g(a)= 0, then the limit is the same as $$lim_{x->a}\frac{\frac{df}{dx}}{\frac{dg}{dx}}$$.

If you get things like $$0^0$$ or $$\infty^{\infty}$$ (as your second limit), you can take logarithms to reduct to the first case.

Homework Helper
mousesgr said:
qs 1 is not 0/0 or inf/inf from
how do consider L'Hopital's rule?
#1 is in form 0 / 0.
Since:
$$\lim_{x \rightarrow 0} (1 + x) ^ {\frac{1}{x}} = e$$
So the numerator will tend to 0 as x approaches 0. The denominator also tends to 0. So it's 0 / 0.
You can use L'Hopital's rule to solve for #1.
-------------------
#2 is $$1 ^ \infty$$
First, you can try to take logs of both sides.
So let $$y = \lim_{x \rightarrow \infty} \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x$$. So:
$$\ln y = \ln \left\{ \lim_{x \rightarrow \infty} \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x \right\} = \lim_{x \rightarrow \infty} \ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x$$
$$= \lim_{x \rightarrow \infty} x \ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] = \lim_{x \rightarrow \infty} \frac{\ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right]}{\frac{1}{x}}$$
This is 0 / 0. So again, you can apply L'Hopital's rule to find the limit.
Viet Dao,