Limit of $f(1-x^2)$: 3, 1, 4, or undefined?

Question

A couple of days ago in AP Calculus, we had an assignment with this problem. Earlier today in class, we received our grades on the assignment... and everyone got this singular problem wrong. However... I don't think she was able to explain it all that well either.

For those who can't see the image for whatever reason, you have a quadratic function that terminates with a hole at $(1,3)$, a point at $(1,2)$, and then a clearly different quadratic function that starts at $(1,4)$. The question is then:

$$ \lim_{x\to 0} f(1-x^2) = ? $$

with the answer choices of $1$,$2$,$3$,$4$, and $\varnothing$. everyone chose either $1$ ($\lim_{x\to 0} 1-x^2 = 1$), $2$ ($f(1) = 2$), or $\varnothing$ ($\lim_{x\to 1} f(x) = \varnothing$). According to the assessment (which was not made by my teacher), the answer was 3.

The way my teacher explained it seems pretty absurd. I don't remember much of it, but she said you had to adjust the values for the functions by flipping it across the x-axis, shifting it up 1, and then finding $lim_{x\to0}$. Most of us are super confused, and since I have an account for this community, I pretty much got nominated to ask for a solution from people significantly more experienced than us.

Send us help on this, please.

Answer 1

Write $y=x^2$. Since $x^2\ge0$, if the limit is rewritten with $y$ then $0$ is approached from the right: $$\lim_{x\to0}f(1-x^2)=\lim_{y\to0^+}f(1-y)$$ Then writing $z=1-y$, $z$ approaches $1$ from the left as the sign is flipped: $$=\lim_{y\to1^-}f(y)=3$$

Answer 2

When $x$ is close to $0$, $1-x^2$ is always close to, but smaller than $1$. Therefore, whatever function $g$ we have, we can say $$\lim_{x\to 0} g(1-x^2) = \lim_{y\nearrow1} g(y)$$ which means, in your question, the limit will be $3$.

Answer 3

Sure, the answer is $3$, $1-x^2<0$ (unless $x=0$, but the value of your function at $0$ doesn't matter), and therefore$$\lim_{x\to0}f(1-x^2)=\lim_{x\to1^-}f(x)=3.$$