$$\lim_{x\to-2}\frac{x^2-16}{x+4}$$
Since it doesn't tell us the limit.. Should i start with something like, claim $\lim f(x)=-6$ and then start the proof as usual? Thanks
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This looks as if you really wanted the limit when $,x\to -4,$ and not when $,x\to -2,$ ... – DonAntonio Apr 28 '13 at 12:21
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Nah, because I have another practise question that is exactly the same thing, but with x→−4 – Apr 28 '13 at 12:30
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HINT: When $x\ne-4$,
$$\frac{x^2-16}{x+4}=x-4\;.$$
Brian M. Scott
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@user74824: In this case you can. However, if you’re required to give an $\epsilon$-$\delta$ proof, you’ll find it easier using $x-4$ than using the original fraction. – Brian M. Scott Apr 28 '13 at 12:37
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@DonAntonio: Amazing how one automatically reads it that way, isn’t it? – Brian M. Scott Apr 28 '13 at 12:47
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I tried letting L=x-4 but everything cancels out... Am I starting this the wrong way with |f(x)-L|<$\epsilon$..? $|\frac{x^2-16}{x+4}-(x-4)|<\epsilon$ – Apr 28 '13 at 12:47
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@user74824: Why would you do that? The limit is a number. What number is it? – Brian M. Scott Apr 28 '13 at 12:48
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@user74824: That’s right. And since the fraction reduces to $x-4$ near $-2$, you’ll find that you can take $\delta=\epsilon$ as long as $\epsilon<2$. – Brian M. Scott Apr 28 '13 at 12:52
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Indeed so, @Brian. Professional deformation, I'd call it, since if $,x\to a\neq -4,$ then the limit is boringly trivial. I think that's what caused me to misread. – DonAntonio Apr 28 '13 at 13:21
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Let $\varepsilon>0$
Then $|f(x)-L|=|\frac{x^2-16}{x+4}-(-6)|=|x-4+6|=|x+2|.....$
Eleven-Eleven
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