Question

evaluate the following: $lim_{x
ightarrow0}\frac{sin x}{x}$
1
$infty$
-1
dne
0
$-infty$
question 11
1 pts
suppose that on the interval $(-2,5)$, we know that $4 - x^{2}leq f(x)leq5 - 2x$. what is $lim_{x
ightarrow1}f(x)$?
-2
0
dne
5
3

Explanation:

Step1: Recall important limit

The limit $\lim_{x
ightarrow0}\frac{\sin x}{x}=1$ is a well - known limit in calculus. It can be proven geometrically or using L'Hopital's rule.

Step2: For the second limit, use the Squeeze Theorem

The Squeeze Theorem states that if $g(x)\leq f(x)\leq h(x)$ for all $x$ in some open interval containing $a$ (except possibly at $x = a$) and $\lim_{x
ightarrow a}g(x)=\lim_{x
ightarrow a}h(x)=L$, then $\lim_{x
ightarrow a}f(x)=L$.
First, find $\lim_{x
ightarrow1}(4 - x^{2})$:
Substitute $x = 1$ into $4 - x^{2}$, we get $4-1^{2}=3$.
Then, find $\lim_{x
ightarrow1}(5 - 2x)$:
Substitute $x = 1$ into $5 - 2x$, we get $5-2\times1 = 3$.
Since $\lim_{x
ightarrow1}(4 - x^{2})=\lim_{x
ightarrow1}(5 - 2x)=3$ and $4 - x^{2}\leq f(x)\leq5 - 2x$, by the Squeeze Theorem, $\lim_{x
ightarrow1}f(x)=3$.

Answer:

1. A. 1
2. E. 3