How to find $\lim_{x\rightarrow \infty }\left ( \frac{x-2}{x-3} \right)^{x}$?

$$\lim_{\infty}(\frac{x-2}{x-3})^x=\lim_{\infty}(1+\frac{1}{x-3})^{x-3}(1+\frac{1}{x-3})^{3}=e$$

$$\frac{x-2}{x-3}=1+\frac1{x-3},$$

then you can substitute $x$ for $x-3$ and

$$\lim_{x\to\infty}\left(1+\frac1x\right)^{x+3}=e.$$

(The $+3$ in the exponent can be ignored as the cube of $1+1/x$ tends to $1$.)

Hint: write your term in the form $$\left(\left(1+\frac{1}{x-3}\right)^{x-3}\right)^{\frac{x}{x-3}}$$