Question

evaluate the limits.
f(x)=\begin{cases}10x + 8&xleq1\\x^{2}-5&x > 1end{cases}
enter dne if the limit does not exist.
a. (lim_{x
ightarrow1^{-}}f(x)=)
b. (lim_{x
ightarrow1^{+}}f(x)=)
c. (lim_{x
ightarrow1}f(x)=)

Explanation:

Step1: Find left - hand limit

For $\lim_{x
ightarrow1^{-}}f(x)$, since $x
ightarrow1^{-}$ means $x < 1$, we use $f(x)=10x + 8$. Substitute $x = 1$ into $10x+8$.
$10\times1 + 8=18$

Step2: Find right - hand limit

For $\lim_{x
ightarrow1^{+}}f(x)$, since $x
ightarrow1^{+}$ means $x>1$, we use $f(x)=x^{2}-5$. Substitute $x = 1$ into $x^{2}-5$.
$1^{2}-5=-4$

Step3: Find overall limit

The overall limit $\lim_{x
ightarrow1}f(x)$ exists if and only if $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)$. Since $18
eq - 4$, $\lim_{x
ightarrow1}f(x)$ does not exist.

Answer:

a. $18$
b. $-4$
c. DNE