Answer to Question #309605 in Calculus for Amira

Question

Activity in Limit Theorems

Directions: Assume the following.

1. lim f(x) = 3/4

x→c

2. lim g(x) = 12

x→c

3. lim h(x) = -3

x→c

Answer 1

Given that

$\lim_{x\rightarrow c}f(x)=\frac{3}{4}$ $\lim_{x\rightarrow c}g(x)=12$ $\lim_{x\rightarrow c}h(x)=-3$

For these values there are no questions, however, a few of the examples are below which will explain how we can use these values using limit theorems.

Ex 1. $\lim_{x\rightarrow c}f(x)+2\lim_{x\rightarrow c}g(x)$

To find $\lim_{x\rightarrow c}f(x)+2\lim_{x\rightarrow c}g(x)$ , we use the given values,

$\lim_{x\rightarrow c}f(x)=\frac{3}{4}$ and $\lim_{x\rightarrow c}g(x)=12$

Therefore,

$=\frac{3}{4}+2(12)=\frac{99}{4}$

Ex 2. $\lim_{x\rightarrow c}f(x)+2\lim_{x\rightarrow c}g(x)$

To find $\lim_{x\rightarrow c}f(x)+2\lim_{x\rightarrow c}g(x)$ , we use the given values,

$\lim_{x\rightarrow c}f(x)=\frac{3}{4}$ and $\lim_{x\rightarrow c}g(x)=12$

Therefore,

$=\frac{3}{4}+2(12)=\frac{99}{4}$

Ex 3. $lim_{x\rightarrow c}\frac{f(x)-3g(x)}{h(x)}$

$=\frac{\lim_{x\rightarrow c}f(x)-3\lim_{x\rightarrow c}g(x)}{\lim_{x\rightarrow c}f(x)}$

$=\frac{\frac{3}{4}-(3)(12)}{-3}$

$=\frac{47}{4}$

Ex 4. $lim_{x\rightarrow c}[h(x)-4f(x)]$

$=lim_{x\rightarrow c}h(x)-4lim_{x\rightarrow c}f(x)$

$=(-3)-4(\frac{3}{4})$

$=-3-3$

$=-6$