$\lim _{x\to \infty \:}|\frac{x+1}{x-2}|^{\sqrt{x^2-4}}$

Question

I need help solving the following tough limit , please.

$$\lim _{x\to \infty} \left|\frac{x+1}{x-2}\right|^{\sqrt{x^2-4}}$$

Answer 1

Since eventually $x-2>0$ we have

$$\lim _{x\to \infty \:}\left|\frac{x+1}{x-2}\right|^{\sqrt{x^2-4}}=\lim _{x\to \infty \:}\left(\frac{x+1}{x-2}\right)^{\sqrt{x^2-4}}$$

then we have

$$\left(\frac{x+1}{x-2}\right)^{\sqrt{x^2-4}}=\left(\frac{x-2+3}{x-2}\right)^{\sqrt{x^2-4}}=\left[\left(1+\frac{3}{x-2}\right)^{\frac{x-2}3}\right]^{\frac{3\sqrt{x^2-4}}{x-2}}\to e^3$$

indeed by $y=\frac{x-2}3 \to \infty$

$$\left(1+\frac{3}{x-2}\right)^{\frac{x-2}3}=\left(1+\frac{1}{y}\right)^{y}\to e$$

and

$$\frac{3\sqrt{x^2-4}}{x-2}=3\cdot \frac x x \frac{\sqrt{1-4/x^2}}{1-2/x}\to 3\cdot 1\cdot 1 =3$$

Answer 2

Write the expression as $e^{\sqrt {x^{2}-4} \log (\frac {x+1} {x-2})}=e^{\sqrt {x^{2}-4} \log (1+\frac 3 {x-2})}$. Use that fact that $\frac {log (1+y)} y \to 1$ as $ y\to 0$. Finally factor $\sqrt {x^{2}-4}$ as $\sqrt {x-2} \sqrt {x+2}$. You will get the limit as $e^{3}$.

Answer 3

$$=\dfrac{\left(\lim_{x\to\infty}\left|1+\dfrac1x\right|^x\right)^{\lim_{x\to\infty}\dfrac{\sqrt{x^2-2}}x}}{\left(\lim_{x\to\infty}\left|1-\dfrac2x\right|^{-x/2}\right)^{\lim_{x\to\infty}\dfrac{-2\sqrt{x^2-2}}x}}=\dfrac{e^{1}}{e^{-2}}=?$$