find each indicated quantity if it exists. let $f(x)=\begin{cases}x^{2},& \text{for }x < - 1\\-2x,&…

find each indicated quantity if it exists. let $f(x)=\begin{cases}x^{2},& \text{for }x < - 1\\-2x,& \text{for }x>-1end{cases}$. complete parts (a) through (d)\n(a) select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. $limlimits_{x\rightarrow - 1^{+}}f(x)=2$ (type an integer.)\nb. the limit does not exist.\n(b) select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. $limlimits_{x\rightarrow - 1^{-}}f(x)=1$ (type an integer.)\nb. the limit does not exist.\n(c) select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. $limlimits_{x\rightarrow - 1}f(x)=square$ (type an integer.)\nb. the limit does not exist.

find each indicated quantity if it exists. let $f(x)=\begin{cases}x^{2},& \text{for }x < - 1\\-2x,& \text{for }x>-1end{cases}$. complete parts (a) through (d)\n(a) select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. $limlimits_{x\rightarrow - 1^{+}}f(x)=2$ (type an integer.)\nb. the limit does not exist.\n(b) select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. $limlimits_{x\rightarrow - 1^{-}}f(x)=1$ (type an integer.)\nb. the limit does not exist.\n(c) select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. $limlimits_{x\rightarrow - 1}f(x)=square$ (type an integer.)\nb. the limit does not exist.

Answer

Explanation:

Step1: Recall right - hand limit definition

For $\lim_{x\rightarrow - 1^{+}}f(x)$, when $x>-1$, $f(x)=-2x$. Substitute $x = - 1$ into $f(x)=-2x$. $f(x)=-2\times(-1)=2$

Step2: Recall left - hand limit definition

For $\lim_{x\rightarrow - 1^{-}}f(x)$, when $x < - 1$, $f(x)=x^{2}$. Substitute $x=-1$ into $f(x)=x^{2}$. $f(x)=(-1)^{2}=1$

Step3: Check overall limit

The overall limit $\lim_{x\rightarrow - 1}f(x)$ exists if and only if $\lim_{x\rightarrow - 1^{-}}f(x)=\lim_{x\rightarrow - 1^{+}}f(x)$. Since $1\neq2$, $\lim_{x\rightarrow - 1}f(x)$ does not exist.

Answer:

(A) A. $\lim_{x\rightarrow - 1^{+}}f(x)=2$ (B) A. $\lim_{x\rightarrow - 1^{-}}f(x)=1$ (C) B. The limit does not exist.