Question

let $lim_{x
ightarrow a}g(x)=3$, $lim_{x
ightarrow a}h(x)=6$, $lim_{x
ightarrow a}f(x)=0$.
find following limits if they exist. if not, enter dne (does not exist) as your answer.

1. $lim_{x

ightarrow a}\frac{f(x)}{g(x)}$

1. $lim_{x

ightarrow a}\frac{g(x)}{h(x)}$

1. $lim_{x

ightarrow a}(h(x))^{-1}$

1. $lim_{x

ightarrow a}g(x)cdot f(x)$

1. $lim_{x

ightarrow a}\frac{g(x)}{f(x)}$

1. $lim_{x

ightarrow a}sqrt{h(x)}$

Explanation:

Step1: Use limit - quotient rule

The limit - quotient rule states that if $\lim_{x
ightarrow a}u(x)=L_1$ and $\lim_{x
ightarrow a}v(x)=L_2
eq0$, then $\lim_{x
ightarrow a}\frac{u(x)}{v(x)}=\frac{L_1}{L_2}$. Given $\lim_{x
ightarrow a}f(x) = 0$ and $\lim_{x
ightarrow a}g(x)=3
eq0$, so $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$.
$\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{0}{3}=0$

Step2: Use limit - quotient rule

Given $\lim_{x
ightarrow a}g(x)=3$ and $\lim_{x
ightarrow a}h(x)=6
eq0$, by the limit - quotient rule $\lim_{x
ightarrow a}\frac{g(x)}{h(x)}=\frac{\lim_{x
ightarrow a}g(x)}{\lim_{x
ightarrow a}h(x)}$.
$\lim_{x
ightarrow a}\frac{g(x)}{h(x)}=\frac{3}{6}=\frac{1}{2}$

Step3: Use limit - power rule

The limit - power rule states that if $\lim_{x
ightarrow a}u(x)=L$, then $\lim_{x
ightarrow a}(u(x))^n = L^n$ for any real number $n$. Given $\lim_{x
ightarrow a}h(x)=6$, and $n=-1$, so $\lim_{x
ightarrow a}(h(x))^{-1}=\frac{1}{\lim_{x
ightarrow a}h(x)}$.
$\lim_{x
ightarrow a}(h(x))^{-1}=\frac{1}{6}$

Step4: Use limit - product rule

The limit - product rule states that if $\lim_{x
ightarrow a}u(x)=L_1$ and $\lim_{x
ightarrow a}v(x)=L_2$, then $\lim_{x
ightarrow a}(u(x)\cdot v(x))=L_1\cdot L_2$. Given $\lim_{x
ightarrow a}g(x)=3$ and $\lim_{x
ightarrow a}f(x)=0$, so $\lim_{x
ightarrow a}(g(x)\cdot f(x))=\lim_{x
ightarrow a}g(x)\cdot\lim_{x
ightarrow a}f(x)$.
$\lim_{x
ightarrow a}(g(x)\cdot f(x))=3\times0 = 0$

Step5: Analyze the limit

Given $\lim_{x
ightarrow a}g(x)=3$ and $\lim_{x
ightarrow a}f(x)=0$, the form is $\frac{\lim_{x
ightarrow a}g(x)}{\lim_{x
ightarrow a}f(x)}=\frac{3}{0}$. Division by zero is undefined, so $\lim_{x
ightarrow a}\frac{g(x)}{f(x)}$ does not exist. So $\lim_{x
ightarrow a}\frac{g(x)}{f(x)}=\text{DNE}$

Step6: Use limit - root rule

The limit - root rule states that if $\lim_{x
ightarrow a}u(x)=L\geq0$ for $n$ even, and $\lim_{x
ightarrow a}u(x)=L$ for $n$ odd, then $\lim_{x
ightarrow a}\sqrt[n]{u(x)}=\sqrt[n]{L}$. Given $\lim_{x
ightarrow a}h(x)=6\geq0$ and $n = 2$, so $\lim_{x
ightarrow a}\sqrt{h(x)}=\sqrt{\lim_{x
ightarrow a}h(x)}$.
$\lim_{x
ightarrow a}\sqrt{h(x)}=\sqrt{6}$

Answer:

1. $0$
2. $\frac{1}{2}$
3. $\frac{1}{6}$
4. $0$
5. DNE
6. $\sqrt{6}$