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

Answer

Given that

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

$\lim_{x\rightarrow c}g(x)=12$ 

$\lim_{x\rightarrow c}h(x)=-3$

We can use these values in the following examples to understand some of the limit theorems.

limit theorems.

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

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

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

Therefore,

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

$=\frac{3}{4}+(12)=\frac{51}{4}$ and $=\frac{3}{4}-(12)=\frac{-45}{4}$

Ex 2. $\lim_{x\rightarrow c}[(2.f(x))+\sqrt{12.g(x)}]$

To find $\lim_{x\rightarrow c}[(2.f(x))+\sqrt{12.g(x)}]$ , we use the given values,

$\lim_{x\rightarrow c}[(2.f(x))+\sqrt{12.g(x)}]\\=[(2.\lim_{x\rightarrow c}f(x))+\sqrt{12.\lim_{x\rightarrow c}g(x)}]$

$=[(2.\frac{4}{3})+\sqrt{(12)(12)}]$

$=\frac{8}{3}+12=\frac{44}{3}$

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

Therefore,

$=2(\frac{3}{4})+(12)=\frac{27}{2}$

Ex 3. $lim_{x\rightarrow c}4\frac{f(x)-g(x)}{2h(x)}$

$=\frac{\lim_{x\rightarrow c}4f(x)-\lim_{x\rightarrow c}g(x)}{\lim_{x\rightarrow c}2h(x)}$

$=\frac{4\lim_{x\rightarrow c}f(x)-\lim_{x\rightarrow c}g(x)}{2\lim_{x\rightarrow c}h(x)}$

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

$=\frac{3}{2}$

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

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

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

$=-3-3$

$=-6$