Q. 87 - Calculus

Question

Prove the second part of Theorem $1.29$: If $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is of the form $\frac{1}{0^{-}}$, then $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=-\infty$.

Step-by-Step Solution

Answer

It is proved that If $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is of the form $\frac{1}{0^{-}}$, then $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=-\infty$.

Step 1. Given Information

We are given two functions $f\left(x\right) \text{and} g\left(x\right)$.

Step 2. Proving the statement

Given $\epsilon >0$, we can choose ${\delta }_1>0$ to get $u\left(x\right)$ within $\epsilon$ of L and also choose ${\delta }_2$ to get $l\left(x\right)$ within $\epsilon$ of L.

If $\delta =\min \left({\delta }_1,{\delta }_2\right)$, then whenever $x\in (c-\delta ,c)\cup (c,c+\delta )$ we also have,

$L-\epsilon <l\left(x\right)\le f\left(x\right)\le u\left(x\right)<L+\epsilon$

Hence, if $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is of the form $\frac{1}{0^{-}}$ then the left part will be the result, and it will be,

$\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=-\infty$.