Question

given that
lim(x→3) f(x) = 9 lim(x→3) g(x) = -2 lim(x→3) h(x) = 0,
find each limit, if it exists. (if an answer does not exist, enter dne.)
(a) lim(x→3) f(x) + 3g(x)
(b) lim(x→3) g(x)^3
(c) lim(x→3) √f(x)
(d) lim(x→3) 4f(x)/g(x)
(e) lim(x→3) g(x)/h(x)
(f) lim(x→3) g(x)h(x)/f(x)

Explanation:

Step1: Use limit - sum rule

$\lim_{x
ightarrow3}[f(x)+3g(x)]=\lim_{x
ightarrow3}f(x)+3\lim_{x
ightarrow3}g(x)$

Step2: Substitute given limits

$=9 + 3\times(-2)=9 - 6 = 3$

Step3: For $\lim_{x

ightarrow3}[g(x)]^{3}$, use power - rule of limits
$\lim_{x
ightarrow3}[g(x)]^{3}=[\lim_{x
ightarrow3}g(x)]^{3}$

Step4: Substitute the value of $\lim_{x

ightarrow3}g(x)$
$=(-2)^{3}=-8$

Step5: For $\lim_{x

ightarrow3}\sqrt{f(x)}$, use root - rule of limits
$\lim_{x
ightarrow3}\sqrt{f(x)}=\sqrt{\lim_{x
ightarrow3}f(x)}$

Step6: Substitute the value of $\lim_{x

ightarrow3}f(x)$
$=\sqrt{9}=3$

Step7: For $\lim_{x

ightarrow3}\frac{4f(x)}{g(x)}$, use quotient - rule of limits
$\lim_{x
ightarrow3}\frac{4f(x)}{g(x)}=\frac{4\lim_{x
ightarrow3}f(x)}{\lim_{x
ightarrow3}g(x)}$

Step8: Substitute the values of $\lim_{x

ightarrow3}f(x)$ and $\lim_{x
ightarrow3}g(x)$
$=\frac{4\times9}{-2}=-18$

Step9: For $\lim_{x

ightarrow3}\frac{g(x)}{h(x)}$, since $\lim_{x
ightarrow3}h(x) = 0$ and $\lim_{x
ightarrow3}g(x)=-2$
The limit is of the form $\frac{-2}{0}$, so $\lim_{x
ightarrow3}\frac{g(x)}{h(x)}=\text{DNE}$

Step10: For $\lim_{x

ightarrow3}\frac{g(x)h(x)}{f(x)}$, use product - and quotient - rules of limits
$\lim_{x
ightarrow3}\frac{g(x)h(x)}{f(x)}=\frac{\lim_{x
ightarrow3}g(x)\cdot\lim_{x
ightarrow3}h(x)}{\lim_{x
ightarrow3}f(x)}$

Step11: Substitute the values of $\lim_{x

ightarrow3}g(x)$, $\lim_{x
ightarrow3}h(x)$ and $\lim_{x
ightarrow3}f(x)$
$=\frac{(-2)\times0}{9}=0$

Answer:

(a) 3
(b) -8
(c) 3
(d) -18
(e) DNE
(f) 0