# Thread: [SOLVED] L'Hopital Rule II

1. ## [SOLVED] L'Hopital Rule II

I was working on another problem, and I get $\displaystyle \frac {1 + \frac{1}{x}}{1}$ and the limit as it approaches x is 0, but apparently the answer here is 2

how? what about $\displaystyle \frac {1}{x}$

2. Originally Posted by >_<SHY_GUY>_<
I was working on another problem, and I get $\displaystyle \frac {1 + \frac{1}{x}}{1}$ and the limit as it approaches x is 0, but apparently the answer here is 2

how? what about $\displaystyle \frac {1}{x}$
what is the original problem?

3. it was $\displaystyle (e^x + x)^\frac{1}{x}$
as x approaches 0

4. Originally Posted by >_<SHY_GUY>_<
it was $\displaystyle (e^x + x)^\frac{1}{x}$
as x approaches 0
let $\displaystyle y = (e^x + x)^{\frac{1}{x}}$

$\displaystyle \ln{y} = \frac{\ln(e^x + x)}{x}$

$\displaystyle \lim_{x \to 0} \frac{\ln(e^x + x)}{x} =$

$\displaystyle \lim_{x \to 0} \frac{e^x + 1}{e^x + x} = 2$

since $\displaystyle \lim_{x \to 0} \ln{y} = 2$ , $\displaystyle \lim_{x \to 0} y = e^2$

5. here's another approach by using elementary limits:

for $\displaystyle x\ne0$ let $\displaystyle \left( e^{x}+x \right)^{\frac{1}{x}}=\exp \left( \frac{\ln \left( 1-\left( 1-\left( e^{x}+x \right) \right) \right)}{1-\left( e^{x}+x \right)}\cdot \frac{1-\left( e^{x}+x \right)}{x} \right),$ thus as $\displaystyle x\longrightarrow0$ this tends to $\displaystyle \exp \left( -1\times \left( -1-1 \right) \right)=e^{2}.$