Question

assignment 2: problem 28 (1 point)
let $f(x)=\begin{cases}-\frac{5}{x + 2},& \text{if }x < - 2\\2x + 5,& \text{if }x > - 2end{cases}$
calculate the following limits. if the limit doesnt exist but it makes sense to call it $infty$ enter infinity, for $-infty$ enter -infinity; in other cases where the limit does not exist enter dne.
$lim_{x
ightarrow - 2^{-}}f(x)=square$
$lim_{x
ightarrow - 2^{+}}f(x)=square$
$lim_{x
ightarrow - 2}f(x)=square$
note: you can earn partial credit on this problem.
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Explanation:

Step1: Left - hand limit

As $x\to - 2^{-}$, $f(x)=-\frac{5}{x + 2}$. Substitute values approaching $-2$ from the left. $\lim_{x\to - 2^{-}}-\frac{5}{x + 2}=\text{Infinity}$.

Step2: Right - hand limit

As $x\to - 2^{+}$, $f(x)=2x + 5$. Substitute $x=-2$ into $2x + 5$, $\lim_{x\to - 2^{+}}(2x + 5)=2\times(-2)+5 = 1$.

Step3: Overall limit

Since $\lim_{x\to - 2^{-}}f(x)
eq\lim_{x\to - 2^{+}}f(x)$, $\lim_{x\to - 2}f(x)=\text{DNE}$.

Answer:

$\lim_{x\to - 2^{-}}f(x)=\text{Infinity}$
$\lim_{x\to - 2^{+}}f(x)=1$
$\lim_{x\to - 2}f(x)=\text{DNE}$