Finding limit of $\left(\frac{n^2 + n}{n^2 + n + 2}\right)^n$

Question

Please, help me to find limit of this sequence:
$\lim_{n\to \infty} \left(\frac{n^2 + n}{n^2 + n + 2}\right)^n$

Answer 1

We have $$ \bigg( \frac{n^{2}+n}{n^{2}+n+2} \bigg)^{n} = \bigg( 1 + \frac{2}{n^{2}+n} \bigg)^{-n} = \exp \bigg[ -n \log \bigg( 1 + \frac{2}{n^{2}+n}\bigg) \bigg] = \exp \bigg( \frac{-2}{n+1} - \frac{2}{n+1}o(1) \bigg) \to e^{0} = 1 $$ as $n \to \infty$.

Answer 2

Hint. Notice that: $$\forall n\in\mathbb{N},\frac{n^2+n}{n^2+n+2}=1-\frac{2}{n^2+n+2}.$$ And now, remind that: $$\forall n\in\mathbb{N},\left(\frac{n^2+n}{n^2+n+2}\right)^n=\exp\left[n\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right].$$

Answer 3

$$\lim_{n\to\infty}\left(\frac{n^2+n}{n^2+n+2}\right)^n=$$ $$\lim_{n\to\infty}\exp\left(\ln\left(\left(\frac{n^2+n}{n^2+n+2}\right)^n\right)\right)=$$ $$\lim_{n\to\infty}\exp\left(n\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)=$$ $$\exp\left(\lim_{n\to\infty}n\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\ln\left(\frac{n^2+n}{n^2+n+2}\right)}{\frac{1}{n}}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\frac{\text{d}}{\text{d}n}\left(\ln\left(\frac{n^2+n}{n^2+n+2}\right)\right)}{\frac{\text{d}}{\text{d}n}\left(\frac{1}{n}\right)}\right)=$$ $$\exp\left(\lim_{n\to\infty}-\frac{2n(2n+1)}{(n+1)(n^2+n+2)}\right)=$$

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Since $-2n(2n+1)$ grows asymptotically slower than $(n+1)(n^2+n+2)$ as $n$ approaches $\infty$:

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$$\exp\left(0\right)=e^0=1$$

Answer 4

Write the expression as

$$\left [ \left ( 1 + \frac{2}{n^2 +n}\right)^{(n^2+n)/2} \right ]^{(2n)/(n^2+n)}.$$

As $n \to \infty,$ the term inside the brackets $\to e,$ and the outer exponent $\to 0.$ The limit is therefore $e^0 = 1.$