Question

given that
lim_{x\to1} f(x)=1 lim_{x\to1} g(x)= - 5 lim_{x\to1} h(x)=0,
find the limits, if they exist. (if an answer does not exist, enter dne.)
(a) lim_{x\to1} f(x)+4g(x)
(b) lim_{x\to1} g(x)^3
(c) lim_{x\to1} \sqrt{f(x)}
(d) lim_{x\to1} \frac{5f(x)}{g(x)}
(e) lim_{x\to1} \frac{g(x)}{h(x)}
(f) lim_{x\to1} \frac{g(x)h(x)}{f(x)}
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Explanation:

Step1: Use limit - sum rule

The limit of a sum $\lim_{x
ightarrow a}[f(x)+g(x)]=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)$ and the constant - multiple rule $\lim_{x
ightarrow a}[cf(x)] = c\lim_{x
ightarrow a}f(x)$.
For $\lim_{x
ightarrow 1}[f(x)+4g(x)]$, we have $\lim_{x
ightarrow 1}[f(x)+4g(x)]=\lim_{x
ightarrow 1}f(x)+4\lim_{x
ightarrow 1}g(x)$.
Substitute $\lim_{x
ightarrow 1}f(x) = 1$ and $\lim_{x
ightarrow 1}g(x)=-5$:
$1 + 4\times(-5)=1-20=-19$.

Step2: Use power - rule for limits

The power - rule states that $\lim_{x
ightarrow a}[f(x)]^n=[\lim_{x
ightarrow a}f(x)]^n$.
For $\lim_{x
ightarrow 1}[g(x)]^3$, substitute $\lim_{x
ightarrow 1}g(x)=-5$:
$(-5)^3=-125$.

Step3: Use root - rule for limits

The root - rule for limits is $\lim_{x
ightarrow a}\sqrt{f(x)}=\sqrt{\lim_{x
ightarrow a}f(x)}$ (when $\lim_{x
ightarrow a}f(x)\geq0$ for even roots).
For $\lim_{x
ightarrow 1}\sqrt{f(x)}$, substitute $\lim_{x
ightarrow 1}f(x) = 1$:
$\sqrt{1}=1$.

Step4: Use quotient - rule for limits

The quotient - rule for limits is $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$ (when $\lim_{x
ightarrow a}g(x)
eq0$).
For $\lim_{x
ightarrow 1}\frac{5f(x)}{g(x)}$, we have $\lim_{x
ightarrow 1}\frac{5f(x)}{g(x)}=\frac{5\lim_{x
ightarrow 1}f(x)}{\lim_{x
ightarrow 1}g(x)}$.
Substitute $\lim_{x
ightarrow 1}f(x) = 1$ and $\lim_{x
ightarrow 1}g(x)=-5$:
$\frac{5\times1}{-5}=-1$.

Step5: Analyze the quotient

For $\lim_{x
ightarrow 1}\frac{g(x)}{h(x)}$, since $\lim_{x
ightarrow 1}g(x)=-5$ and $\lim_{x
ightarrow 1}h(x)=0$, the limit is of the form $\frac{-5}{0}$, so $\lim_{x
ightarrow 1}\frac{g(x)}{h(x)}=\text{DNE}$.

Step6: Use product - rule for limits

The product - rule for limits is $\lim_{x
ightarrow a}[f(x)g(x)]=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)$.
For $\lim_{x
ightarrow 1}\frac{g(x)h(x)}{f(x)}$, we have $\lim_{x
ightarrow 1}\frac{g(x)h(x)}{f(x)}=\frac{\lim_{x
ightarrow 1}g(x)\cdot\lim_{x
ightarrow 1}h(x)}{\lim_{x
ightarrow 1}f(x)}$.
Substitute $\lim_{x
ightarrow 1}f(x) = 1$, $\lim_{x
ightarrow 1}g(x)=-5$ and $\lim_{x
ightarrow 1}h(x)=0$:
$\frac{-5\times0}{1}=0$.

Answer:

(a) - 19
(b) - 125
(c) 1
(d) - 1
(e) DNE
(f) 0