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I don't understand how the limit does not exist for the composite function. The limit as x approaches -2 for g(x) is zero. So, the last step is to evaluate h(0), which is -1. Yes, there is a hole at x=0 but that doesn't mean you can't evaluate h(0).

enter image description here

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  • $\begingroup$ From the JPEG I see that there are two values for $h(0)$, no? and you must consider the limit to these values of $g(-2)$ $\endgroup$ – higgs May 30 '19 at 8:48
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    $\begingroup$ You do not evaluate $h(0)$. You should determine $\lim_{x\to0}h(x)$ instead. $\endgroup$ – user1551 May 30 '19 at 8:48
  • $\begingroup$ What is $h(g(-2.001))$? $h(g(-1.999))$? Does this gap close towards $0$ as $x \to -2$? $\endgroup$ – Henry May 30 '19 at 9:02
  • $\begingroup$ You do evaluate h(0) I thought. I am following the method I see all over the internet which is finding the limit of the inner function (in this case g(x)) and then plugging this value in for the outer function (or h(x)). We don't care about limits for h(x) only g(x). $\endgroup$ – user532874 May 30 '19 at 9:02
  • $\begingroup$ limit of inner function, plug into outer function. That works only if the outer function is continuous. The reason you see it all over the internet, is that functions all over the internet are continuous. $\endgroup$ – GEdgar May 30 '19 at 12:41
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It's true that $\lim_{x\to-2}g(x)=0$, but as $g(x)$ goes to $0$, it gets there from two directions. Namely, from above and from below. It goes through values like $-0.1$, $-0.001$, $-0.0001$, etc. from below and through values like $0.1$, $0.001$, $0.0001$, etc. from above. That's equivalent to evaluating $\lim_{x\to 0}h(x)$ which, according to the theory of limits, is really two one-sided limits under the hood. And what does the function $h(x)$ approach as you go to $0$ from the right and from the left?

$$\lim_{x\to 0^-}h(x)=1$$ and $$\lim_{x\to 0^+}h(x)=-1.$$

Those two limits don't agree and thus the limit itself does not exist.

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  • $\begingroup$ I get that but the formula for getting the limit of h(g(x)) or any composite function for that matter only involves finding the limit of the inner function (in this case g(x)) and then plugging this value in for h(x). We don't care about limits for h(x) only g(x). $\endgroup$ – user532874 May 30 '19 at 9:00
  • $\begingroup$ As you're going to $−2$, the expression $g(x)$ is becoming a quantity that more and more resembles $0$. So, as you go to $−2$, you get values like $0.001$, $−0.001$, $0.00001$, $−0.00001$, etc. This is equivalent to evaluating $\lim_{x\to0}h(x)$ which does not exist because, from the left and from the right, $h(x)$ approaches different function values. $\endgroup$ – Michael Rybkin May 30 '19 at 9:27
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For the limit to exist the following must hold: $$\lim_{x\uparrow-2}h(g(x))=\lim_{x\downarrow-2}h(g(x))=h(g(0)),$$ Looking at the picture we can see that the left limit $$\lim_{x\uparrow-2}h(g(x))=\lim_{x\uparrow0}h(x)=1$$ the right limit $$\lim_{x\downarrow-2}h(g(x))=\lim_{x\downarrow0}h(x)=-1$$ and $h(g(-2))=h(3)=1$.

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    $\begingroup$ In the first line, only the first equality must hold for the limit of $h\circ g$ to exist. The second need not hold unless we are trying to prove continuity of $h\circ g$. $\endgroup$ – robjohn May 30 '19 at 21:32
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The limit exists if every point near $-2$ in the domain maps to a points near to one another in the image.

If $x$ is slightly greater than $-2$ then $g(x)$ is slighty greater than $0$, and $(h\circ g)(x)$ is slightly greater than $-1$

but if $x$ is slightly less than $-2$ then $g(x)<0$ and $(h\circ g)(x) > 1$

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