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

Question

How to find the limit :

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

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What is the approach for $1^\infty$ ?

Answer 1

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

Answer 2

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

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 $1$.)

Answer 3

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

Answer 4

$$\left(\dfrac{x-2}{x-3}\right)^x=\left(\dfrac{1-\dfrac2x}{1-\dfrac3x}\right)^x$$

Now $$\lim_{x\to\infty}\left(1-\dfrac ax\right)^x=\left[\lim_{x\to\infty}\left(1-\dfrac ax\right)^{-x/a}\right]^{-a}=e^{-a}$$