calculus - The limit of this function as x approaches 2 from the positive direction (the right).

$$\lim_{x \rightarrow 2}\frac {\sqrt{x^2-4}}{x-2}$$

Am I correct in thinking the limit does not exist? Since as x approaches 2 from the right the function increases to infinity and a limit cannot equal infinity. Thanks!

Answer

$$\lim_{x\rightarrow 2^+} \frac{\sqrt{x^2-4}}{x-2}=\lim_{x\rightarrow 2^+}\frac{\sqrt{(x-2)(x+2)}}{x-2}=\lim_{x\rightarrow 2^+}\frac{\sqrt{(x-2)}\sqrt{(x+2)}}{x-2}=\lim_{x\rightarrow 2^+}\frac{\sqrt{(x+2)}}{\sqrt{(x-2)}}$$

Then as you said the denominator tends to $0^+$ and hence the whole limit tends to $+\infty$