# Find asymptote of $f(x)=(1+x)^{1/x}$

Find asymptote of $f(x)=(1+x)^{1/x}$

I am thinking, maybe I differentiate and let $f'(x)=0$? But how do I even differentiate the equation?

Or maybe I do $\lim_{n\to\infty} (1+x)^{1/x}$ ... but I dont really see how do I do this?

I think the answer is $y=1$? I need a better proof tho?

-
You might also want to look at the behavior as $x\to-1^{+}$... – J. M. May 1 '12 at 8:38

There is no $n$; you want to find $\lim\limits_{x\to\infty}(1+x)^{1/x}$. This is a so-called $\infty^0$ indeterminate form, and there is a standard technique for computing such limits.
Let $$L=\lim_{x\to\infty}(1+x)^{1/x}\;,$$ and take logarithms to get $$\ln L=\ln\lim_{x\to\infty}(1+x)^{1/x}=\lim_{x\to\infty}\ln(1+x)^{1/x}=\lim_{x\to\infty}\frac1x\ln(1+x)=\lim_{x\to\infty}\frac{\ln(1+x)}x\;,$$ where the second equality is true because the logarithm is a continuous function. This last limit is an $\frac{\infty}{\infty}$ form, so you can use l'Hospital's rule to evaluate it. Once you have $\ln L$, remember that $L=e^{\ln L}$.