Question

f(x) =
−x, x ≤ −1
1 − x^2, −1 < x < 1
x − 1, x ≥ 1
.
Evaulate each of the following or explain why it does not exist.

lim
x→−1 f(x),
lim
x→0 f(x),
lim
x→1 f(x) .

Answer 1

\[find~\lim_{x \rightarrow-1 -}f(x)~and ~\lim_{x \rightarrow-1+ }f(x)\]
similarly
find
\[\lim_{x \rightarrow 1-}f(x)~and~\lim_{x \rightarrow 1+}f(x)\]
similarly for x=0

Answer 2

Yes! Haha, my prof didn't post solutions so I was wondering if I got the answers right. lim of -1 I got DNE. For lim of 0 I got 1. And for lim of 1 I got 0.

Answer 3

\[\lim_{x \rightarrow -1-}f(x)=\lim_{x \rightarrow -1-}\left( -x \right)\]
put x=1-h,h>0
\[h \rightarrow0~as~x \rightarrow-1-\]
\[\lim_{x \rightarrow -1-}f(x)=\lim_{h \rightarrow 0}\left\{ -\left( -1-h \right) \right\}=\lim_{h \rightarrow 0}\left( 1+h \right)=1\]
\[\lim_{x \rightarrow -1+}f(x)=\lim_{x \rightarrow -1+}\left( 1-x^2 \right)=1-1=0\]
\[\lim_{x \rightarrow -1-}f(x) \neq \lim_{x \rightarrow -1+}f(x)\]
\[Hence \lim_{x \rightarrow -1}f(x) ~does~\not~ exist.\]
so you are correct.
similarly your other limits are also correct.