Question

for the function shown below, find (if the quantity exists) (a) $lim_{x
ightarrow0^{+}}f(x)$, (b) $lim_{x
ightarrow0^{-}}f(x)$, (c) $lim_{x
ightarrow0}f(x)$, and (d) $f(0)$.
$f(x)=\begin{cases}10 - x^{2},& \text{for }xleq0\\10 + x^{2},& \text{for }x > 0end{cases}$
(b) select the correct choice below and fill in any answer boxes in your choice.
a. $lim_{x
ightarrow0^{-}}f(x)=$
b. the limit does not exist.

Explanation:

Step1: Find the right - hand limit ($x\to0^{+}$)

For $x > 0$, $f(x)=10 + x^{2}$. So, $\lim_{x\to0^{+}}f(x)=\lim_{x\to0^{+}}(10 + x^{2})$. Substitute $x = 0$ into $10 + x^{2}$, we get $10+0^{2}=10$.

Step2: Find the left - hand limit ($x\to0^{-}$)

For $x\leq0$, $f(x)=10 - x^{2}$. So, $\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}(10 - x^{2})$. Substitute $x = 0$ into $10 - x^{2}$, we get $10-0^{2}=10$.

Step3: Find the overall limit ($x\to0$)

Since $\lim_{x\to0^{+}}f(x)=\lim_{x\to0^{-}}f(x)=10$, then $\lim_{x\to0}f(x)=10$.

Step4: Find $f(0)$

For $x = 0$, we use the part of the piece - wise function where $x\leq0$, so $f(0)=10-0^{2}=10$.

Answer:

(A) $\lim_{x\to0^{+}}f(x)=10$
(B) $\lim_{x\to0^{-}}f(x)=10$
(C) $\lim_{x\to0}f(x)=10$
(D) $f(0)=10$