Question 207395
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*[tex \Large \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 2^+}\ \LARGE \left(\frac{\sqrt{x-2}}{x^2-4}\right)]


Since both numerator and denominator go to zero as *[tex \Large x] approaches 2 from the right, you need to use L'Hôpital's Rule.


If *[tex \Large \lim_{x\rightarrow c}f(x)\ =\ \lim_{x\rightarrow c}g(x)\ =\ 0] or *[tex \Large \pm\infty] and if *[tex \Large \lim_{x\rightarrow c}f'(x)/g'(x)] exists, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow c}\frac{f(x)}{g(x)}\ =\ \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)]


The rule also applies to one-sided limits, so:


If *[tex \Large \lim_{x\rightarrow c^\pm}f(x)\ =\ \lim_{x\rightarrow c^\pm}g(x)\ =\ 0] or *[tex \Large \pm\infty] and if *[tex \Large \lim_{x\rightarrow c^\pm}f'(x)/g'(x)] exists, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow c^\pm}\frac{f(x)}{g(x)}\ =\ \lim_{x\rightarrow c^\pm}\frac{f'(x)}{g'(x)]


So take the first derivative of each of the numerator and denominator functions.  


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 2^+}\left(\frac{\sqrt{x-2}}{x^2-4}\right)\ =\ \lim_{x\rightarrow 2^+}\left(\frac{\frac{1}{2\sqrt{x-2}}}{2x}\right)\ =\ +\infty]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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