Evaluation of $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^{\sqrt{n}}$ Can anyone help me in solving the limit problem:
$$\lim_\limits{n \to \infty} \left(1 + \frac{1}{n}\right)^{\sqrt{n}}$$
 A: There’s a standard trick for dealing with such limits. Let $$y=\left(1+\frac1n\right)^{\sqrt{n}}\;.$$
Then
$$\ln y=\sqrt{n}\ln\left(1+\frac1n\right)=\frac{\ln\left(1+\frac1n\right)}{n^{-1/2}}\;.$$
The log is continuous, so $\lim\limits_{n\to\infty}\ln y=\ln\lim\limits_{n\to\infty}y$, and therefore
$$\lim_{n\to\infty}y=e^{\lim\limits_{n\to\infty}\ln y}\;.$$
Now use l’Hospital’s rule to evaluate $\lim\limits_{n\to\infty}\ln y$.
(In this problem one can actually avoid these calculations by making use of the fact that
$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\;,$$
but the general method is worth knowing.)
A: $$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{\sqrt{n}}=\lim_{n \to \infty}\left (1 + \frac{1}{n}\right)^{n\cdot\frac{\sqrt{n}}{n}}=e^{\lim_{n\to\infty}\frac{\sqrt n}{n}}=e^0=1$$
A: You can have a look at this alternative approach. I asked it ago and got @Brian's concrete answer. This is the link. According to it you would have $$\lim_{n\to\infty}\left(1+1/n\right)^{\sqrt{n}}=\exp(k)$$ wherein $$k=\lim_{n\to +\infty}\big(1+1/n-1\big)\sqrt{n}=0$$
A: $$\left(1+\frac{1}{n}\right)^{\sqrt{n}}=\left[\left(1+\frac{1}{n}\right)^n\right]^{\sqrt{n}/n}$$
A: $$
\begin{array}{l}
 y = \left( {1 + \frac{1}{n}} \right)^{\sqrt n }  \Leftrightarrow y = {\mathop{\rm e}\nolimits} ^{\sqrt n \ln \left( {1 + \frac{1}{n}} \right)}  \\ 
 \mathop {\lim }\limits_{n \to  + \infty } \left( {1 + \frac{1}{n}} \right)^{\sqrt n }  = \mathop {\lim }\limits_{n \to  + \infty } {\mathop{\rm e}\nolimits} ^{\sqrt n \ln \left( {1 + \frac{1}{n}} \right)}  = {\mathop{\rm e}\nolimits} ^{\mathop {\lim }\limits_{n \to  + \infty } \sqrt n \ln \left( {1 + \frac{1}{n}} \right)}  = e^0  = 1 \\ 
 \end{array}
$$
