1. ## L'Hospital's Rule

Hi, i'm trying to help my sibling with these math problems, but i'm having some trouble:

1. lim x-> -infinity (x^2)^(e^x)

2. and lim x -> 0 (1/x-1/ln(x+1))

For one 1. I use the natural log to make e^x * ln(x^2) and i've tried put both over eachother but it goes nowhere. e^x will continue to be zero. I've used the rule repeatedly and cannot find an answer. I don't see part of the derivative of ln(x^2) or 1/ln(x^2) helping yet either.

For two i've put both over a common denominator but still am having trouble getting an answer

thanks

2. Hello,
Originally Posted by jarny
Hi, i'm trying to help my sibling with these math problems, but i'm having some trouble:

1. lim x-> -infinity (x^2)^(e^x)

2. and lim x -> 0 (1/x-1/ln(x+1))

For one 1. I use the natural log to make e^x * ln(x^2) and i've tried put both over eachother but it goes nowhere. e^x will continue to be zero. I've used the rule repeatedly and cannot find an answer. I don't see part of the derivative of ln(x^2) or 1/ln(x^2) helping yet either.

$\displaystyle e^x \ln(x^2)=\frac{\ln(x^2)}{\frac{1}{e^x}}=\frac{\ln( x^2)}{e^{-x}}$

as x goes to - infinity, x^2 goes to + infinity and hence ln(x^2) goes to + infinity
as x goes to - infinity, -x goes to + infinity and hence $\displaystyle e^{-x}$ goes to + infinity.

So now you can apply the rule. Just differentiate (by using chain rule) and see if it can help.

For two i've put both over a common denominator but still am having trouble getting an answer
I guess it's better you show what you've done =)

3. Hello, jarny!

I think I got #2 . . .

$\displaystyle 2)\;\;\lim_{x\to0}\left[\frac{1}{x}-\frac{1}{\ln(x+1)}\right]$
Combine the fractions: .$\displaystyle \frac{\ln(x+1)-x}{x\ln(x+1)}$

Since this goes to $\displaystyle \frac{0}{0}$. we can apply L'Hopital.

. . Then we have: .$\displaystyle \frac{\frac{1}{x+1} - 1}{x\cdot\frac{1}{x+1} + \ln(x+1)}$

Multiply by $\displaystyle \frac{x+1}{x+1}\!:\quad \frac{1 - (x+1)}{x + (x+1)\ln(x+1)} \;=\;\frac{-x}{x+(x+1)\ln(x+1)}$

Since this goes to $\displaystyle \frac{0}{0}$, we can apply L'Hopital again.

. . Then we have: .$\displaystyle \frac{-1}{1 + (x+1)\cdot\frac{1}{x+1} + \ln(x+1)} \;=\;\frac{-1}{2 + \ln(x+1)}$

Therefore: .$\displaystyle \lim_{x\to0}\left[\frac{-1}{2+\ln(x+1)}\right] \;=\;\frac{-1}{2+0} \;=\;\boxed{-\frac{1}{2}}$