How to resolve $\lim_{x\to\infty} (1+\frac{1}{x^n})^x$

Question

So I was interested in limits of the following form

$$\lim_{x\to\infty} (1+\frac{1}{x^n})^x \qquad n\geqslant1$$

when $n=1$, the limit is $e$, of course. I was then able to prove the limit approaches $1$ when $n\geqslant2$. I cannot figure out how deal with the case $1\lt n \lt 2$. I know it converges, but I can't figure out where it converges.

Answer 1

When $n=1$ it is easy to do!

Because $$(1+\frac{1}{x^n})^x =\left((1+\frac{1}{x^n})^{x^n}\right)^{\frac{1}{x^{n-1}}},n>1.$$ And use the result:If $a_n\to a>0$ and $b_n\to b$, then $a_{n}^{b_n}\to a^b.$

In this case, let $a_n=(1+\frac{1}{x^n})^{x^n}$ and $b_n=\frac{1}{x^{n-1}}$, so $a_n\to e$ and $b_n\to 0$, and the desired limit is $1$.

Answer 2

Consider $$a= \left(1+\frac{1}{x^n}\right)^x\implies \log(a)=x \log \left(1+\frac{1}{x^n}\right)$$ Since $x$ is large, using equivalents $$\log(a) \sim \frac{1}{x^{n-1}}$$ which goes to $0$ if $n>1$. So $a \to 1$.