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).
[Math] Why does limit not exist
calculuslimits
Related Solutions
Note: $\delta q$ is just one variable, not two. (Just incase that is confusing.)
Hey there, these are all great answers. But I'd like to be a bit more precise about what's confusing you, if I can:
If I have a function f and that function is not defined at some x, then asking for the derivative of the function at x makes no sense since there is no f(x) at x.
I think you're way too used to the variables and not seeing the entire point: When we take a derivative of a function (let's call it $z(q)$), we're taking a limit that eventually simplifies:
$ \frac{d}{dq} z(q)=\lim_{\delta q \to 0} \frac{z(q+\delta q)-z(q)}{\delta q} $
Now, I'm sure you're aware of that and all, but it seems you've confused your terms. You have used the term "$x$" in two different ways without realizing it. Think about your statement rewritten:
If I have a function $z(q)$ and that function is not defined at some point, then asking for the derivative of the function at that point makes no sense since there is no $z(q)$ at that point.
Notice how I didn't confuse the variable of the function with the point? Now, even that statement isn't entirely accurate. The question becomes: What are you precisely thinking? Consider: We have $z(q)$ and, let's presume, its derivative: $z'(q)$. Now, are you saying: Why should we consider $z'(x)$ (a specific value) when $z'(q)$ (a specific function) is discontinuous at $x$ (a specific point)? That makes perfect sense. Do you see the difference and how this clarifies precisely what you're saying and gets rid of the confusion? I think you understand this matter really well, you've just confused yourself by using the same thing to denote very different things.
I can't find the derivative at that point since it doesn't exist, But the limit is 2. But the limit is the derivative, and the derivative doesn't exist! I'm confused.
You've got a precise example of what I'm saying. You're confusing the limit of the derivative at the point with the limit of the function. Let me illustrate: Your derivative of $z$ is a specific limit. But it is NOT the same limit that you take when you take the limit of $z'$ at a point that $z'$ is undefined at. More annoyingly stated: $z'(q)=\lim_{\delta q \to 0}\frac{z(q+\delta q)-z(q)}{\delta q}$ whereas (the second) is $\lim_{q \to y}z'(q)=\lim_{q \to y}\lim_{\delta q \to 0}\frac{z(q+\delta q)-z(q)}{\delta q}$ (where $y$ is your undefined point).
I'm sorry if anything I've said here is useless or redundant, I just hope this helps. This is my first answer on math, so if I'm made any horrendous faux pas, I'm sorry.
Try the paths
- $y=x$ with $x>0$; and
- $y=-x$ with $x>0$.
Both paths get arbitrarily close to the origin. Compute the limit restricted to each of those paths (you may need to apply L'Hôpital's rule). If you get two different results, the limit does not exist (because the limit can't depend on the path).
Best Answer
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.