Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$

Question

How can I find this limit?

$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$

Firstly I thought I can use the limit:

$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$

by adding $1$ and subtracting $1$ from the original limit. However, since $3-2x$ $\rightarrow - \infty$ and not $+\infty$, I got nowhere. Then I tried finding the logarithm of this limit. It resulted in a $\dfrac{0}{0}$ indeterminate form, I tried L'Hospital, but again, it led me nowhere. Either I made some mistakes in the calculations, or I should use a different approach.

Answer 1

Yes, add and subtract $1$. You will get

$$\left[ \left( 1+\frac{\sqrt{x^2+2x-3}-x-2}{x+2} \right)^{\frac{x+2}{\sqrt{x^2+2x-3}-x-2}} \right]^{\frac{(\sqrt{x^2+2x-3}-x-2)(3-2x)}{x+2}}$$

The part inside the $\left[...\right]$ tends to $e$.

Then compute the limit of the exponent

$$\begin{align}\frac{(\sqrt{x^2+2x-3}-x-2)(3-2x)}{x+2}&=\frac{((x^2+2x-3)-(x+2)^2)(3-2x)}{(\sqrt{x^2+2x-3}+x+2)(x+2)}\\ &=\frac{(-2x-7)(3-2x)}{(\sqrt{x^2+2x-3}+x+2)(x+2)}\\ &=\frac{(-2-7/x)(3/x-2)}{(\sqrt{1+2/x-3/x^2}+1+2/x)(1+2/x)}\\ &\to2\end{align}$$

Therefore, the original limit is $e^{2}$

Answer 2

\begin{align*} &\left(\dfrac{\sqrt{x^{2}+2x-3}}{x+2}\right)^{3-2x}\\ &=\left(\dfrac{\sqrt{x^{2}+2x-3}}{\sqrt{x^{2}+2x}}\right)^{3-2x}\left(\dfrac{\sqrt{x^{2}+2x}}{x+2}\right)^{3-2x}\\ &=\left(1-\dfrac{3}{x^{2}+2x}\right)^{3/2-x}\left(\dfrac{x}{x+2}\right)^{3/2-x}\\ &=\left(1-\dfrac{3}{x^{2}+2x}\right)^{-(x^{2}+2x)(x-3/2)/(x^{2}+2x)}\left(1-\dfrac{2}{x+2}\right)^{-(x+2)(x-3/2)/(x+2)}\\ &\rightarrow 1\cdot e^{2}\\ &=e^{2}. \end{align*}

Answer 3

Let $y =\bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$. Then,

$$\ln y = (3-2x)\ln \sqrt{\dfrac{x^2+2x-3}{(x+2)^2}} $$ $$= \frac{3-2x}2 \ln \left(1-\frac2{x+2}+O(\frac1{(x+2)^2})\right) = \frac{\ln\left( 1-\frac2{x+2}\right)+O(\frac1{(x+2)^2})}{\frac1{-(x+2)}+O(\frac1{(x+2)^2})}$$

Therefore,

$$\lim\limits_{x \to \infty} \ln y = \lim\limits_{x \to \infty}\frac{\ln\left( 1-\frac2{x+2}\right)}{\frac1{-(x+2)}} =2$$

which leads to $\lim\limits_{x \to \infty} y = e^2$.

Answer 4

Take logarithm and substitute by $x=1/y$ to get $$ \frac{{\left(3 \, y - 2\right)} \log\left(3 \, y + 1\right) - 2 \, {\left(3 \, y - 2\right)} \log\left(2 \, y + 1\right) + 3 \, y \log\left(-y + 1\right) - 2 \, \log\left(-y + 1\right)}{2 \, y} = 2+O(y) $$ as $y \to 0$, where we use $\log(1+y)=y+O(y^2)$.