Question

find each indicated quantity if it exists. let $f(x)=\

$$\begin{cases}x^{2},&\\text{for }x < - 1\\\\-2x,&\\text{for }x > - 1\\end{cases}$$

$. complete parts (a) through (d)

(b) select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. $\lim_{x\to - 1^{+}}f(x)=1$ (type an integer.)
b. the limit does not exist.

(c) select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. $\lim_{x\to - 1}f(x)=$ (type an integer.)
b. the limit does not exist.

(d) select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. $f(-1)=$ (type an integer.)
b. the function is not defined at $x = - 1$.

Explanation:

Step1: Find the left - hand limit as $x\to - 1^{-}$

For $x < - 1$, $f(x)=x^{2}$. So, $\lim_{x\to - 1^{-}}f(x)=\lim_{x\to - 1^{-}}x^{2}=(-1)^{2}=1$.

Step2: Find the right - hand limit as $x\to - 1^{+}$

For $x > - 1$, $f(x)=-2x$. So, $\lim_{x\to - 1^{+}}f(x)=\lim_{x\to - 1^{+}}(-2x)=(-2)\times(-1) = 2$.

Step3: Determine the overall limit as $x\to - 1$

Since $\lim_{x\to - 1^{-}}f(x)=1$ and $\lim_{x\to - 1^{+}}f(x)=2$, and $1
eq2$, $\lim_{x\to - 1}f(x)$ does not exist.

Step4: Check the value of the function at $x = - 1$

The function $f(x)$ is defined as $x^{2}$ for $x < - 1$ and $-2x$ for $x > - 1$. There is no definition for $x=-1$. So, $f(-1)$ is not defined.

Answer:

(B) A. $\lim_{x\to - 1^{-}}f(x)=1$
(C) B. The limit does not exist.
(D) B. The function is not defined at $x = - 1$.