Evaluating $\lim_{n \to \infty} (n^2+1)^{n/(n^2+1)}$

Question

I can't figure out how to solve this limit:

$$\lim_{n \to \infty} (n^2+1)^{n/(n^2+1)}$$

I can calculate the limit of the power which is $0$, but then appears the indetermination ($\infty^0$).

Can someone help me, please?

Answer 1

If you know $n^{1/n}\to 1,$ you can say

$$1\le (n^2+1)^{n/(n^2+1)} \le (2n^2)^{1/n}=2^{1/n}(n^{1/n})^2 \to 1\cdot 1^2 =1.$$

Answer 2

HINT

If your limit exists and equals $L$, then $$ \ln L = \ln \left( \lim_{n \to \infty} (n^2+1)^\frac{n}{n^2+1}\right) = \lim_{n \to \infty} \ln \left( (n^2+1)^\frac{n}{n^2+1}\right) $$ Can you finish?

UPDATE

$$ \begin{split} \ln L &= \ln \left( \lim_{n \to \infty} (n^2+1)^\frac{n}{n^2+1}\right) \\ &= \lim_{n \to \infty} \ln \left( (n^2+1)^\frac{n}{n^2+1}\right) \\ &= \lim_{n \to \infty} \frac{n}{n^2+1} \ln \left(n^2+1\right) \\ \end{split} $$ Note that asymptotically $n^2 +1 \approx n^2$ so you have $$ \frac{n \ln \left(n^2+1\right)}{n^2+1} = n \times \frac{\ln \left(n^2+1\right)}{\ln\left(n^2\right)} \times \frac{n^2}{n^2+1} \times \frac{\ln\left(n^2\right)}{n^2} \approx \frac{n \ln \left(n^2\right)}{n^2} = \frac{2n \ln n}{n^2} = 0, $$ (which can be formalized by noting that the middle two fractions both converge to $1$ as $n \to \infty$).

Can you find $L$ now?

Answer 3

By continuity of $x\mapsto e^x$ and $x\mapsto\log x$ we have $$ \lim_{n\to\infty} (n^2+1)^{\frac n{n^2+1}} = e^{\lim_{n\to\infty} \frac{n\log(n^2+1)}{n^2+1}} = e^0 = 1. $$