Question

if $lim_{x
ightarrow1}\frac{f(x)-3}{x - 1}=2$, what is $lim_{x
ightarrow1}f(x)$?
3
2
1
0
dne
question 9
1 pts
evaluate the following:
$lim_{x
ightarrowinfty}\frac{sin x}{x}$
$infty$
-1
$-infty$
0
1
dne

Explanation:

Step1: Analyze first limit

Given $\lim_{x
ightarrow1}\frac{f(x)-3}{x - 1}=2$. Since the denominator $x - 1
ightarrow0$ as $x
ightarrow1$, for the limit of the fraction to exist, the numerator $f(x)-3$ must also approach $0$ as $x
ightarrow1$. So $\lim_{x
ightarrow1}(f(x)-3)=0$.

Step2: Solve for $\lim_{x

ightarrow1}f(x)$
Using the limit - difference rule $\lim_{x
ightarrow a}(u(x)-v(x))=\lim_{x
ightarrow a}u(x)-\lim_{x
ightarrow a}v(x)$. We have $\lim_{x
ightarrow1}(f(x)-3)=\lim_{x
ightarrow1}f(x)-\lim_{x
ightarrow1}3$. Since $\lim_{x
ightarrow1}3 = 3$ and $\lim_{x
ightarrow1}(f(x)-3)=0$, then $\lim_{x
ightarrow1}f(x)-3 = 0$, so $\lim_{x
ightarrow1}f(x)=3$.

Step3: Analyze second limit

For $\lim_{x
ightarrow\infty}\frac{\sin x}{x}$, we know that $- 1\leqslant\sin x\leqslant1$ for all $x$. Then $-\frac{1}{x}\leqslant\frac{\sin x}{x}\leqslant\frac{1}{x}$.

Step4: Apply Squeeze Theorem

We know that $\lim_{x
ightarrow\infty}-\frac{1}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{1}{x}=0$. By the Squeeze Theorem, $\lim_{x
ightarrow\infty}\frac{\sin x}{x}=0$.

Answer:

Question 1: A. 3
Question 2: D. 0