Question

name: natalie arellano
period: 5
ap calculus ab: limits quiz #1

1. if $f(x)=\begin{cases}x^{2}-10, & x < - 4\\-3 - 2x, & xgeq - 4end{cases}$

a. $lim_{x
ightarrow - 4^{+}}f(x)=$
b. $lim_{x
ightarrow - 4}f(x)=$
c. $lim_{x
ightarrow - 5}f(x)=$

Explanation:

Step1: Find $\lim_{x

ightarrow - 4^{+}}f(x)$
Since $x
ightarrow - 4^{+}$ means $x>-4$, we use $f(x)=-3 - 2x$. Substitute $x = - 4$ into $-3 - 2x$: $-3-2\times(-4)=-3 + 8=5$.

Step2: Find $\lim_{x

ightarrow - 4^{-}}f(x)$
Since $x
ightarrow - 4^{-}$ means $x<-4$, we use $f(x)=x^{2}-10$. Substitute $x=-4$ into $x^{2}-10$: $(-4)^{2}-10=16 - 10 = 6$.
Since $\lim_{x
ightarrow - 4^{+}}f(x)
eq\lim_{x
ightarrow - 4^{-}}f(x)$, $\lim_{x
ightarrow - 4}f(x)$ does not exist.

Step3: Find $\lim_{x

ightarrow - 5}f(x)$
Since $x
ightarrow - 5$ means $x<-4$, we use $f(x)=x^{2}-10$. Substitute $x=-5$ into $x^{2}-10$: $(-5)^{2}-10=25 - 10=15$.

Answer:

a. $\lim_{x
ightarrow - 4^{+}}f(x)=5$
b. $\lim_{x
ightarrow - 4}f(x)$ does not exist
c. $\lim_{x
ightarrow - 5}f(x)=15$