Q. 88 - Calculus

Question

Prove the second part of Theorem $1.30$: If $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is of the form $\frac{1}{-\infty }$, then $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=0$.

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2. Proving the statement

Consider a function $f\left(x\right)$ approaches $1$ that is $f\left(x\right)\to 1$.

For any $\epsilon >0$, the function $f\left(x\right)$ satisfies that $1-\epsilon <f\left(x\right)<1+\epsilon$.

Consider a function $g\left(x\right)$ approaches $-\infty$ that is $g\left(x\right)\to -\infty$.

Consider a number M which is less than $0$ that is $M<0$. Therefore, the function $g\left(x\right)$ can be written as $g\left(x\right)<M$.

The number can be written as $M=\frac{-2}{\epsilon }$ and consider $\epsilon =1$.

Divide the function $f\left(x\right)$ by the function $g\left(x\right)$,

$$\begin{gathered} \frac{1+\epsilon }{M}<\frac{f\left(x\right)}{g\left(x\right)}<1-\epsilon \\ \frac{1+\epsilon }{\left(-\frac{2}{\epsilon }\right)}<\frac{f\left(x\right)}{g\left(x\right)}<1-\epsilon \\ \frac{1+1}{\left(-\frac{2}{1}\right)}<\frac{f\left(x\right)}{g\left(x\right)}<1-1 \\ -1<\frac{f\left(x\right)}{g\left(x\right)}<0 \end{gathered}$$

Hence, if $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is in the form of $\frac{1}{-\infty }$ then $\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=0$. Hence the given statement is proved.