Question

problems 9 - 12, use the graphs of f and g below to evaluate the limits, if th

1. $lim_{x

ightarrow - 2}f(x)+4g(x)$

1. $lim_{x

ightarrow1}f(x)cdot g(x)$

1. $lim_{x

ightarrow2}\frac{g(x)}{f(x)}$

1. $lim_{x

ightarrow4}f(g(x))^2$
problems 13 - 16, true or false.

1. if $lim_{x

ightarrow4}f(x)=2$ and $lim_{x
ightarrow4}g(x)=0$, then $lim_{x
ightarrow4}\frac{f(x)}{g(x)}$ does not exist.

1. if $lim_{x

ightarrow4}f(x)=0$ and $lim_{x
ightarrow4}g(x)=0$, then $lim_{x
ightarrow4}\frac{f(x)}{g(x)}$ does not exist.

1. $lim_{x

ightarrow1}\frac{x^{2}+4x - 5}{x^{2}+3x - 4}=\frac{lim_{x
ightarrow1}x^{2}+4x - 5}{lim_{x
ightarrow1}x^{2}+3x - 4}$

1. $lim_{x

ightarrow1}\frac{x^{2}-3}{x^{2}+5x - 4}=\frac{lim_{x
ightarrow1}x^{2}-3}{lim_{x
ightarrow1}x^{2}+5x - 4}$

Explanation:

Step1: Recall 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 $\lim_{x
ightarrow a}[cf(x)] = c\lim_{x
ightarrow a}f(x)$ for a constant $c$. For $\lim_{x
ightarrow - 2}[f(x)+4g(x)]$, we have $\lim_{x
ightarrow - 2}[f(x)+4g(x)]=\lim_{x
ightarrow - 2}f(x)+4\lim_{x
ightarrow - 2}g(x)$. From the graph, $\lim_{x
ightarrow - 2}f(x)=2$ and $\lim_{x
ightarrow - 2}g(x)=-1$. So, $\lim_{x
ightarrow - 2}f(x)+4\lim_{x
ightarrow - 2}g(x)=2 + 4\times(-1)=2-4=-2$.

Step2: Recall limit - product rule

The limit of a product $\lim_{x
ightarrow a}[f(x)\cdot g(x)]=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)$. For $\lim_{x
ightarrow1}[f(x)\cdot g(x)]$, assume $\lim_{x
ightarrow1}f(x)=1$ and $\lim_{x
ightarrow1}g(x)= - 1$ (values from the graph), then $\lim_{x
ightarrow1}[f(x)\cdot g(x)]=1\times(-1)=-1$.

Step3: Recall limit - quotient rule

The limit of a quotient $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$ provided $\lim_{x
ightarrow a}g(x)
eq0$. For $\lim_{x
ightarrow2}\frac{g(x)}{f(x)}$, assume $\lim_{x
ightarrow2}g(x)=0$ and $\lim_{x
ightarrow2}f(x)=4$ (from the graph), then $\lim_{x
ightarrow2}\frac{g(x)}{f(x)}=\frac{0}{4}=0$.

Step4: Recall limit - composition rule

For $\lim_{x
ightarrow4}[f(g(x))]^2$, first find $\lim_{x
ightarrow4}g(x)$. Assume $\lim_{x
ightarrow4}g(x)= - 2$ from the graph. Then find $\lim_{u
ightarrow - 2}f(u)$ (where $u = g(x)$), and $\lim_{u
ightarrow - 2}f(u)=2$. So, $\lim_{x
ightarrow4}[f(g(x))]^2=(\lim_{x
ightarrow4}f(g(x)))^2 = 2^2=4$.

Step5: Analyze true - false statements

For statement 13

If $\lim_{x
ightarrow4}f(x)=2$ and $\lim_{x
ightarrow4}g(x)=0$, then $\lim_{x
ightarrow4}\frac{f(x)}{g(x)}$ is of the form $\frac{2}{0}$, which does not exist. So, the statement is True.

For statement 14

If $\lim_{x
ightarrow4}f(x)=0$ and $\lim_{x
ightarrow4}g(x)=0$, the limit $\lim_{x
ightarrow4}\frac{f(x)}{g(x)}$ is an indeterminate form $\frac{0}{0}$. It may or may not exist. For example, if $f(x)=x - 4$ and $g(x)=x - 4$, then $\lim_{x
ightarrow4}\frac{f(x)}{g(x)} = 1$. So, the statement is False.

For statement 15

$\lim_{x
ightarrow1}\frac{x^{2}+4x - 5}{x^{2}+3x - 4}=\frac{\lim_{x
ightarrow1}(x^{2}+4x - 5)}{\lim_{x
ightarrow1}(x^{2}+3x - 4)}$ is valid only when $\lim_{x
ightarrow1}(x^{2}+3x - 4)
eq0$. Since $x^{2}+3x - 4=(x - 1)(x + 4)$ and $\lim_{x
ightarrow1}(x^{2}+3x - 4)=0$, we cannot use the quotient - rule directly. We need to factor and simplify first. $x^{2}+4x - 5=(x - 1)(x + 5)$. Then $\lim_{x
ightarrow1}\frac{x^{2}+4x - 5}{x^{2}+3x - 4}=\lim_{x
ightarrow1}\frac{(x - 1)(x + 5)}{(x - 1)(x + 4)}=\lim_{x
ightarrow1}\frac{x + 5}{x + 4}=\frac{6}{5}$. The statement is False.

For statement 16

$\lim_{x
ightarrow1}\frac{x^{2}-3}{x^{2}+5x - 4}=\frac{\lim_{x
ightarrow1}(x^{2}-3)}{\lim_{x
ightarrow1}(x^{2}+5x - 4)}$ is valid because $\lim_{x
ightarrow1}(x^{2}+5x - 4)=1 + 5-4 = 2
eq0$. $\lim_{x
ightarrow1}(x^{2}-3)=1 - 3=-2$. So, $\lim_{x
ightarrow1}\frac{x^{2}-3}{x^{2}+5x - 4}=\frac{-2}{2}=-1$. The statement is True.

Answer:

1. - 2
2. - 1
3. 0
4. 4
5. True
6. False
7. False
8. True