Question

let
f(x)=\begin{cases}5x^{2}+2x, & x < 1\\3 - x, & xgeq1end{cases}
what can be said about the limit
lim_{x
ightarrow1^{-}}\frac{f(x)+2}{2 + f(1 + x)}?
the limit exists and equals to 3.
the limit does not exist.
the limit exists and equals to 1.
the limit exists and equals to 9/4.
the limit exists and equals to 7.

Explanation:

Step1: Find $f(x)$ as $x\to 1^{-}$

When $x\to 1^{-}$, $f(x)=5x^{2}+2x$. Substitute $x = 1$ into $f(x)$: $f(1^{-})=5\times1^{2}+2\times1=7$.

Step2: Find $f(1 + x)$ as $x\to 1^{-}$

When $x\to 1^{-}$, $1 + x>1$, so $f(1 + x)=3-(1 + x)=2 - x$. As $x\to 1^{-}$, $\lim_{x\to 1^{-}}f(1 + x)=2-1 = 1$.

Step3: Calculate the limit

Substitute the values of $f(x)$ and $f(1 + x)$ into the limit $\lim_{x\to 1^{-}}\frac{f(x)+2}{2 + f(1 + x)}$. We get $\frac{7 + 2}{2+1}=\frac{9}{3}=3$.

Answer:

The limit exists and equals to 3.