How to find the limit of $\sqrt[^n]{n+1}$ as n approaches $\infty$? My teacher hasn't taught L'Hopital's Rule yet, so I am only able to solve this using some standard limits.
The similar standard limit I can think of is $\lim_{n\to\infty} (1+\frac{a}{n})^n = e^a$.
What I have done so far:
$\lim_{n\to\infty} \sqrt[^n]{n+1}$
$ = \lim_{n\to\infty} (n+1)^{\frac{1}{n}}$
Let $ m = \frac{1}{n} \therefore n = \frac{1}{m}$. Since $n\rightarrow\infty$, $m\rightarrow0$.
$\lim_{m\to0} (1+\frac{1}{m})^m$
But I'm stuck in this step because I don't know how to work out the answer. If I just substitute $0$ in m, $1+\frac{1}{m}$ is undefined. If I rewrite this as $e^{log((1+\frac{1}{m})^m)}$, it's also undefined. I know the limit is $1$ by looking at the graph. I'm wondering how I'm able to solve it in a more rigorous way.
Best Answer
From copper.hat's hint: \begin{equation} \begin{split} \lim_{n\to\infty} \sqrt[^n]{n+1}&=\lim_{n\to\infty} (n+1)^{\frac{1}{n}}\\ &= \lim_{n\to\infty} e^{\log((n+1)^{\frac{1}{n}})}\\ &= \lim_{n\to\infty} e^{({\frac{1}{n}\log(n+1)})}\\ \end{split} \end{equation}
Let $t = \log(n+1)$. We have that
\begin{equation} \begin{split} &e^t = e^{\log(n+1)}=n+1\\ &n = e^t - 1\\ \end{split} \end{equation}
Therefore, our limit equals
$$\displaystyle \lim_{t\to\infty} e^{\frac{t}{e^t-1}}= \large{e}^{\displaystyle \small{\lim_{t\to\infty} \frac{t}{e^t-1}}} = e^0 = 1$$
The limit of the exponent is $0$ because the denominator $e^t$ grows much faster than the numerator $t$.
Another solution:
\begin{equation} \begin{split} \lim_{n\to\infty} \sqrt[^n]{n+1}&=\lim_{n\to\infty} (n+1)^{\frac{1}{n}}\\ &= \lim_{n\to\infty} e^{\log((n+1)^{\frac{1}{n}})}\\ &= \lim_{n\to\infty} e^{({\frac{log(n+1)}{n}})}\ \end{split} \end{equation}
The limit of the exponent is $0$ because the denominator $n$ grows much faster than the numerator $log(n+1)$.