# Evaluation of $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^{\sqrt{n}}$

Can anyone help me in solving the limit problem:

$$\lim_\limits{n \to \infty} \left(1 + \frac{1}{n}\right)^{\sqrt{n}}$$

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.)

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{\sqrt{n}}=\lim_{n \to \infty}\left (1 + \frac{1}{n}\right)^{n\cdot\frac{\sqrt{n}}{n}}=e^{\lim_{n\to\infty}\frac{\sqrt n}{n}}=e^0=1$$

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 $$k=\lim_{n\to +\infty}\big(1+1/n-1\big)\sqrt{n}=0$$

$$\left(1+\frac{1}{n}\right)^{\sqrt{n}}=\left[\left(1+\frac{1}{n}\right)^n\right]^{\sqrt{n}/n}$$

$$\begin{array}{l} y = \left( {1 + \frac{1}{n}} \right)^{\sqrt n } \Leftrightarrow y = {\mathop{\rm e}\nolimits} ^{\sqrt n \ln \left( {1 + \frac{1}{n}} \right)} \\ \mathop {\lim }\limits_{n \to + \infty } \left( {1 + \frac{1}{n}} \right)^{\sqrt n } = \mathop {\lim }\limits_{n \to + \infty } {\mathop{\rm e}\nolimits} ^{\sqrt n \ln \left( {1 + \frac{1}{n}} \right)} = {\mathop{\rm e}\nolimits} ^{\mathop {\lim }\limits_{n \to + \infty } \sqrt n \ln \left( {1 + \frac{1}{n}} \right)} = e^0 = 1 \\ \end{array}$$