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$$f(x) = \frac{\sin{(x-2)}}{x^2-4}$$

I can find right limit in $\pm\infty$ and in $-2$, but I can't find it in $x = 2$. According to my computation it should be infinity, but in the fact it is $\frac{1}{4}$.

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3 Answers 3

up vote 6 down vote accepted

Simply $$\frac{\sin(x-2)}{x^2-4}=\frac{\sin(x-2)}{x-2}\frac{1}{x+2}\to 1\cdot \frac{1}{4}=\frac{1}{4}$$ as $x\to 2$. You just have to prove $$\frac{\sin(x-2)}{x-2}\to 1$$ as $x\to 2$ which follows since $\frac{\sin(x)}{x}\to 1$ as $x\to 0$.

Indeed, as $x\to 2$, $u=x-2\to 0$ and so $$\lim_{x\to 2}\frac{\sin(x-2)}{x-2}=\lim_{u\to 0}\frac{\sin(u)}{u}= 1$$

Alternatively, you can use De l'Hôpital's Rule: $$\lim_{x\to 2}\frac{\sin(x-2)}{x^2-4}\stackrel{l'H}{=}\lim_{x\to 2}\frac{\cos(x-2)}{2x}=\frac{1}{4}$$

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Knowing that $\,\frac{\sin t}{t}\xrightarrow [t\to 0]{}1\,$ , we get

$$\frac{\sin(x-2)}{x^2-4}=\frac{1}{x+2}\frac{\sin(x-2)}{x-2}\xrightarrow [x\to 2]{}\frac{1}{4}\cdot 1=\frac{1}{4}$$

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A parentheses is needed above. –  Babak S. Dec 28 '12 at 11:57
    
Indeed it was. Thanks, Babak. –  DonAntonio Dec 28 '12 at 12:00

Another approach is by using L'Hopital rule. Since both $\lim_{x\to 2}{\sin(x-2)} \to 0$ and $\lim_{x\to2}{x^2-4}\to0$ , by the L'Hopital Rule , the limit becomes $$\lim_{x\to2}{\frac{\cos(x-2)}{2x}} = \frac{1}{4}$$

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See here for formatting math equations. And welcome! –  Daryl Dec 28 '12 at 12:11

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