Question

1. find a so that the function is continuous at every point. guessing a will not earn you points. show your work! $f(x)=\

$$\begin{cases}5 - x, & \\text{if }x < 2 \\\\ ax + 3, & \\text{if }x\\geq2\\end{cases}$$

$ 23. find the limit. you must show how the limit is found using techniques learned in class. no shortcuts will earn points! $\lim_{x\to0}\frac{\sqrt{x^{2}+9}-3}{x^{2}}$ 24. if $3 - 2x^{2}\leq g(x)\leq3\cos x$ for all x, find the limit $\lim_{x\to0}g(x)$. (hint: use the sandwich theorem.)

Explanation:

22.

Step1: Recall continuity condition

For a function to be continuous at $x = 2$, $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)=f(2)$. First, find $\lim_{x
ightarrow2^{-}}f(x)$.
$\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{-}}(5 - x)=5-2 = 3$.

Step2: Find $\lim_{x

ightarrow2^{+}}f(x)$
$\lim_{x
ightarrow2^{+}}f(x)=\lim_{x
ightarrow2^{+}}(ax + 3)=2a+3$.

Step3: Set the two - sided limits equal

Since the function is continuous at $x = 2$, we set $2a+3 = 3$.
Subtract 3 from both sides: $2a=0$.
Divide both sides by 2: $a = 0$.

Step1: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator $\sqrt{x^{2}+9}+3$.
\[

$$\begin{align*} \lim_{x ightarrow0}\frac{\sqrt{x^{2}+9}-3}{x^{2}}&=\lim_{x ightarrow0}\frac{(\sqrt{x^{2}+9}-3)(\sqrt{x^{2}+9}+3)}{x^{2}(\sqrt{x^{2}+9}+3)}\\ &=\lim_{x ightarrow0}\frac{(x^{2}+9)-9}{x^{2}(\sqrt{x^{2}+9}+3)}\\ &=\lim_{x ightarrow0}\frac{x^{2}}{x^{2}(\sqrt{x^{2}+9}+3)} \end{align*}$$

\]

Step2: Simplify the expression

Cancel out $x^{2}$ terms: $\lim_{x
ightarrow0}\frac{1}{\sqrt{x^{2}+9}+3}$.

Step3: Evaluate the limit

Substitute $x = 0$ into the expression: $\frac{1}{\sqrt{0 + 9}+3}=\frac{1}{3 + 3}=\frac{1}{6}$.

Step1: Find $\lim_{x

ightarrow0}(3 - 2x^{2})$
$\lim_{x
ightarrow0}(3 - 2x^{2})=3-2\times0^{2}=3$.

Step2: Find $\lim_{x

ightarrow0}(3\cos x)$
$\lim_{x
ightarrow0}(3\cos x)=3\cos(0)=3\times1 = 3$.

Step3: Apply the sandwich theorem

Since $3 - 2x^{2}\leq g(x)\leq3\cos x$ for all $x$ and $\lim_{x
ightarrow0}(3 - 2x^{2})=\lim_{x
ightarrow0}(3\cos x)=3$, by the sandwich theorem, $\lim_{x
ightarrow0}g(x)=3$.

Answer:

$a = 0$

23.