Question

suppose
\\(\lim_{x\to a}h(x)=1, \lim_{x\to a}f(x)=1, \lim_{x\to a}g(x)=0\\).
find following limits if they exist. enter dne if the limit does not exist.

1. \\(\lim_{x\to a}h(x)+f(x)\\)
2. \\(\lim_{x\to a}h(x)-f(x)\\)
3. \\(\lim_{x\to a}h(x)\cdot g(x)\\)
4. \\(\lim_{x\to a}\frac{h(x)}{f(x)}\\)
5. \\(\lim_{x\to a}\frac{h(x)}{g(x)}\\)
6. \\(\lim_{x\to a}\frac{g(x)}{h(x)}\\)
7. \\(\lim_{x\to a}(f(x))^{2}\\)
8. \\(\lim_{x\to a}\frac{1}{f(x)}\\)
9. \\(\lim_{x\to a}\frac{1}{f(x)-g(x)}\\)

Explanation:

Step1: Use limit - sum rule

$\lim_{x
ightarrow a}(h(x)+f(x))=\lim_{x
ightarrow a}h(x)+\lim_{x
ightarrow a}f(x)=1 + 1=2$

Step2: Use limit - difference rule

$\lim_{x
ightarrow a}(h(x)-f(x))=\lim_{x
ightarrow a}h(x)-\lim_{x
ightarrow a}f(x)=1 - 1=0$

Step3: Use limit - product rule

$\lim_{x
ightarrow a}(h(x)\cdot g(x))=\lim_{x
ightarrow a}h(x)\cdot\lim_{x
ightarrow a}g(x)=1\times0 = 0$

Step4: Use limit - quotient rule

$\lim_{x
ightarrow a}\frac{h(x)}{f(x)}=\frac{\lim_{x
ightarrow a}h(x)}{\lim_{x
ightarrow a}f(x)}=\frac{1}{1}=1$

Step5: Since $\lim_{x

ightarrow a}g(x)=0$ and $\lim_{x
ightarrow a}h(x)=1$
$\lim_{x
ightarrow a}\frac{h(x)}{g(x)}=\text{DNE}$

Step6: Use limit - quotient rule

$\lim_{x
ightarrow a}\frac{g(x)}{h(x)}=\frac{\lim_{x
ightarrow a}g(x)}{\lim_{x
ightarrow a}h(x)}=\frac{0}{1}=0$

Step7: Use limit - power rule

$\lim_{x
ightarrow a}(f(x))^{2}=(\lim_{x
ightarrow a}f(x))^{2}=1^{2}=1$

Step8: Use limit - reciprocal rule

$\lim_{x
ightarrow a}\frac{1}{f(x)}=\frac{1}{\lim_{x
ightarrow a}f(x)}=\frac{1}{1}=1$

Step9: First, find $\lim_{x

ightarrow a}(f(x)-g(x))$
$\lim_{x
ightarrow a}(f(x)-g(x))=\lim_{x
ightarrow a}f(x)-\lim_{x
ightarrow a}g(x)=1-0 = 1$, then $\lim_{x
ightarrow a}\frac{1}{f(x)-g(x)}=\frac{1}{1}=1$

Answer:

1. 2
2. 0
3. 0
4. 1
5. DNE
6. 0
7. 1
8. 1
9. 1