Calculate the limit of $\left(\frac{n+1}{n-5}\right)^{5n-2}$

Question

Calculate $$\lim_{n\to \infty}d_n=\lim_{n\to \infty} \left(\frac{n+1}{n-5}\right)^{5n-2}$$ My problem probably because I think this limit it kind of $1^\infty$ limit and I know it's undefined.

Do I need to use this assumption to find this limit?

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

Answer 1

You can rewrite $$ \frac{n+1}{n-5} = 1+\frac{6}{n-5} $$ from which $$\begin{align} \left(\frac{n+1}{n-5}\right)^{5n-2} &= \left(1+\frac{6}{n-5} \right)^{5n-2} = \left(1+\frac{6}{n-5} \right)^{5(n-5)+23}\\ &= \left(\left(1+\frac{6}{n-5} \right)^{n-5}\right)^{5}\cdot \left(1+\frac{6}{n-5}\right)^{23} \end{align}$$ Now, use the limit you know to show that the first factor will converge to $(e^{6})^5=e^{30}$; while the second is easily seen to converge to $1^{23}=1$.

Answer 2

To simplify we can let $m=n-5\to \infty$

$$\lim_{n \to \infty} \left(\frac{n+1}{n-5}\right)^{5n-2}=\lim_{m \to \infty} \left(\frac{m+6}{m}\right)^{5m+23}=\lim_{m \to \infty} \left[\left(1+\frac{6}{m}\right)^{m}\right]^{\frac{5m+23}{m}}=e^{30}$$