Mathematics Asked by Outsider on February 18, 2021
According to the answers in my book e = 4 and b = -2, I am unsure about e, but for b I would have assumed it to be -1.
Can someone explain to me why the answers are what they are?
For $xto2$, $f(x)$ tends to $-2$ on the left and $+2$ on the right, hence the square tends to $4$ on both sides, and this is the limit (the value at $x=2$ is irrelevant).
Then the limit of $f$ at $x=0$ is $2$, and more precisely $2^-$ because the values remain smaller than $2$. As seen above, the limit of $f$ on the left is $-2$.
Correct answer by Yves Daoust on February 18, 2021
(e): we have $lim_{x to 2-0}f(x)=-2$, hence $lim_{x to 2-0}f(x)^2=4$.
Furthermore: $lim_{x to 2+0}f(x)=2$, hence $lim_{x to 2+0}f(x)^2=4$.
This gives: $lim_{x to 2}f(x)^2=4$.
(b) we have $lim_{x to 0}f(x)=2$ and $f(x)<2$ for $x in (-1,1) setminus {0}.$
Hence $lim_{x to 0}f(f(x))= lim_{z to 2-0}f(z)=-2.$
Answered by Fred on February 18, 2021
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