name: natalie arellano\nperiod: 5\nap calculus ab: limits quiz #1\n1. if $f(x)=\begin{cases}x^{2}-10, & x <…

name: natalie arellano\nperiod: 5\nap calculus ab: limits quiz #1\n1. if $f(x)=\begin{cases}x^{2}-10, & x < - 4\\-3 - 2x, & xgeq - 4end{cases}$\na. $lim_{x\rightarrow - 4^{+}}f(x)=$\nb. $lim_{x\rightarrow - 4}f(x)=$\nc. $lim_{x\rightarrow - 5}f(x)=$

name: natalie arellano\nperiod: 5\nap calculus ab: limits quiz #1\n1. if $f(x)=\begin{cases}x^{2}-10, & x < - 4\\-3 - 2x, & xgeq - 4end{cases}$\na. $lim_{x\rightarrow - 4^{+}}f(x)=$\nb. $lim_{x\rightarrow - 4}f(x)=$\nc. $lim_{x\rightarrow - 5}f(x)=$

Answer

Explanation:

Step1: Find $\lim_{x\rightarrow - 4^{+}}f(x)$

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

Step2: Find $\lim_{x\rightarrow - 4^{-}}f(x)$

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

Step3: Find $\lim_{x\rightarrow - 5}f(x)$

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