Evaluate $\lim_{x\rightarrow \infty}(1+\frac{1}{\sqrt{x}})^{\sqrt{x}}$. Euler's Limit

Question

Evaluate $\lim\limits_{x\rightarrow \infty}(1+\frac{1}{\sqrt{x}})^{\sqrt{x}}$.

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Can I get some help? I am thinking that the limit does not exist. If you approach it from the left and then from the right, I think that the limits do not equal each other. I also suspect that we are dealing with Euler's limit, i.e.

The limits $\lim\limits_{x\rightarrow \infty +} (1+\frac{1}{x})^x$ and $\lim\limits_{x\rightarrow \infty -} (1+\frac{1}{x})^x$ exist and both equal $e$

Otherwise, I have little intuition to go from.

Answer 1

When $\;x\to\infty\;$ we can assume $\;x>0\;$ when doing the limit, so now simply make a substitution:

$$x\leftrightarrow y^2\;\implies\;\;x\to\infty\iff y\to\infty$$

and your limit becomes

$$\lim_{y\to\infty}\left(1+\frac1y\right)^y=e$$

For negative $\;x$'s $\;\sqrt x\;$ isn't defined and thus also the limit of your expression isn't when $\;x\to -\infty\;$

Answer 2

Note that we can write

\begin{equation*} \lim_{x\to\infty} e^{(\sqrt{x}\ln(1+\frac{1}{\sqrt{x}}))} = e^{(\lim_{x\to\infty} \sqrt{x}\ln(1+\frac{1}{\sqrt{x}}))} =e^{\left(\lim_{x\to\infty}\frac{\ln(1+\frac{1}{\sqrt{x}})}{\frac{1}{\sqrt{x}}}\right)}. \end{equation*}

Applying L'Hopital's rule rule gives

\begin{equation*} e^{\left(\lim_{x\to\infty}\frac{1}{1+\frac{1}{\sqrt{x}}}\right)}=e. \end{equation*}