2
$\begingroup$

Can anyone help me in solving the limit problem:

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

$\endgroup$
6
$\begingroup$

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.)

$\endgroup$
4
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.