Solving a limit: $\lim_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x-4}$

Question

Question:

$$\lim_{x \to \infty} \bigg(\frac{x+6}{x+1}\bigg)^{x-4}$$

Attempt: It is quite obvious that the term inside the brackets tends to $1$ while the exponent tends to $\infty$. How would I solve this further?

Answer 1

Hint: Think of $$\lim_{x\to\infty}\left(1+\frac1x\right)^{x}=e$$

Answer 2

HINT:

$$\bigg(\frac{x+6}{x+1}\bigg)^{x-4}=\left(\left(1+\dfrac5{x+1}\right)^{\frac{x+5}5}\right)^5\cdot\left(1+\dfrac5{x+1}\right)^{-9}$$ and use $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$

Answer 3

$$\lim_{x \to \infty} \bigg(\frac{x+6}{x+1}\bigg)^{x-4}=\lim_{x \to \infty} \bigg(\frac{x+1+5}{x+1}\bigg)^{x-4}=\lim_{x \to \infty} \bigg(1+\frac{5}{x+1}\bigg)^{\frac{x+1}{5}((x-4) \frac{5}{x+1})}=e^5$$

Answer 4

Let's get you started on your way.

$$ \begin{align} \lim_{x \to \infty} \bigg(\frac{x+6}{x+1}\bigg)^{x-4}&=\lim_{x \to \infty} \exp\bigg(\log\bigg(\bigg(\frac{x+6}{x+1}\bigg)^{x-4}\bigg)\bigg)\\ &=\lim_{x \to \infty} \exp\bigg((x-4)\log\bigg(\frac{x+6}{x+1}\bigg)\bigg)\\ &=\exp\bigg(\lim_{x \to \infty} (x-4)\log\bigg(\frac{x+6}{x+1}\bigg)\bigg)\\ &=\exp\bigg(\lim_{x \to \infty} \frac{\log\bigg(\frac{x+6}{x+1}\bigg)}{\frac{1}{x-4}}\bigg)\\ \end{align} $$

From here apply l'Hôpital's rule and you should be well on your way to solving it.

Answer 5

Let $$A=\bigg(\frac{x+6}{x+1}\bigg)^{x-4}$$ $$\log(A)=(x-4)\log\bigg(\frac{x+6}{x+1}\bigg)=(x-4)\log\bigg(1+\frac{5}{x+1}\bigg)$$ Now, using that, for small values of $y$ $$\log(1+y)=y-\frac{y^2}{2}+O\left(y^3\right)$$ replace $y$ by $\frac{5}{x+1}$ and get $$\log(A)=5 \frac {x-4} {x+1}-\frac{25}2 \frac {x-4} {(x+1)^2}+\cdots$$

I am sure that you can take from here.