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) $f(0)$.
$f(x)=\begin{cases}15 - x^{2},& \text{for }xleq0\\15 + x^{2},& \text{for }x>0end{cases}$
(a) select the correct choice below and fill in any answer boxes in your choice.
a. $lim_{x
ightarrow0^{+}}f(x)=square$
b. the limit does not exist.

Explanation:

Step1: Recall right - hand limit definition

For $\lim_{x
ightarrow0^{+}}f(x)$, we use the part of the piece - wise function where $x > 0$. Since $f(x)=15 + x^{2}$ for $x>0$.

Step2: Substitute $x = 0$ into the function

Substitute $x = 0$ into $y = 15+x^{2}$. We get $\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}(15 + x^{2})=15+0^{2}$.

Step3: Calculate the limit value

$15+0^{2}=15$.

For $\lim_{x
ightarrow0^{-}}f(x)$, we use the part of the piece - wise function where $x\leq0$. Since $f(x)=15 - x^{2}$ for $x\leq0$. Substitute $x = 0$ into $y = 15 - x^{2}$, we get $\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{-}}(15 - x^{2})=15-0^{2}=15$.

Since $\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{-}}f(x)=15$, then $\lim_{x
ightarrow0}f(x)=15$.

For $f(0)$, we use the part of the piece - wise function where $x\leq0$. Since $f(x)=15 - x^{2}$ for $x\leq0$, then $f(0)=15-0^{2}=15$.

(A)

Answer:

A. $\lim_{x
ightarrow0^{+}}f(x)=15$
(B) $\lim_{x
ightarrow0^{-}}f(x)=15$
(C) $\lim_{x
ightarrow0}f(x)=15$
(D) $f(0)=15$