1. ## Limit

Hello, I need some help for two horrific limits...
$\displaystyle (e ^{(x-1)^2+(1/(x+1))} - (e)^{1/2}) / (ln(x+1) - ln(2))$
when x -> 1.

The whole "(x-1)^2+(1/(x+1))" goes with the "e^"

And, secondly:

$\displaystyle x * ln (x / (x+2))$
when x-> +oo

Thank you!

2. Originally Posted by Gogol
Hello, I need some help for two horrific limits...
$\displaystyle (e ^{(x-1)^2+(1/(x+1))} - (e)^{1/2}) / (ln(x+1) - ln(2))$
when x -> 1.

The whole "(x-1)^2+(1/(x+1))" goes with the "e^"

Thank you!
Since we have a case of $\displaystyle \frac{0}{0}$, L'Hopital's Rule works.

$\displaystyle \frac{d}{dx}\left[\frac{e^{(x-1)^2+(x+1)^{-1}}-\sqrt{e}}{\ln(x+1)-\ln(2)}\right]=\frac{e^{(x-1)^2+(x+1)^{-1}}\cdot[2(x-1)-(x+1)^{-2}]}{(x+1)^{-1}}$

Plug in $\displaystyle x=1$:

$\displaystyle \frac{e^{(1-1)^2+(1+1)^{-1}}\cdot[2(1-1)-(1+1)^{-2}]}{(1+1)^{-1}}=\frac{e^{1/2}\cdot-\frac{1}{4}}{\frac{1}{2}}=\boxed{-\frac{\sqrt{e}}{2}}$

3. Originally Posted by Gogol
Hello, I need some help for two horrific limits...

And, secondly:

$\displaystyle x * ln (x / (x+2))$
when x-> +oo

Thank you!
$\displaystyle \lim_{x\to\infty}x\ln\left(\frac{x}{x+2}\right)=\l im_{x\to\infty}\frac{\ln\left(\frac{x}{x+2}\right) }{\frac{1}{x}}$

Now we have another $\displaystyle \frac{0}{0}$ case, so we can use L'Hopital's Rule again. Taking the derivatives of the top and bottom gives us:

$\displaystyle \frac{\frac{x+2}{x}\cdot\frac{2}{(x+2)^2}}{-\frac{1}{x^2}}=-\frac{2x}{x+2}=-\frac{2x}{x(1+2/x)}=-\frac{2}{1+2/x}$

Now take the limit:

$\displaystyle \lim_{x\to\infty}-\frac{2}{1+2/x}=-\frac{2}{1+0}=\boxed{-2}$

4. Thank you!