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  • If  <img alt="\lim _{x \rightarrow p} f(x)=+\infty \text { where } f(x)=\frac{x^2+a x+m^2}{b x^2+a x+n^2}, b \neq 0 \quad \text {, then } \lim _{x \rightarrow p} \frac{x^2+\frac{a}{b} x+\frac{

If  \lim _{x \rightarrow p} f(x)=+\infty \text { where } f(x)=\frac{x^2+a x+m^2}{b x^2+a x+n^2}, b \neq 0 \quad \text {, then } \lim _{x \rightarrow p} \frac{x^2+\frac{a}{b} x+\frac{n^2}{b}}{\frac{x^2}{b}+x+\frac{m^2}{b}}=\text { ? }
 

Option: 1

+\infty


Option: 2

-\infty


Option: 3

0


Option: 4

\pm \infty


Answers (1)

best_answer

The following property of limits is important.

When the limitis defined as \lim_{x\rightarrow 1}f\left ( x \right )= +\infty, then \lim_{x\rightarrow 1}\frac{1}{f\left ( x \right )}= 0.
The provided function is f(x)=\frac{x^2+a x+m^2}{b x^2+a x+n^2}.
So, the function \frac{1}{f\left ( x \right )} is as followa:

\frac{1}{f\left ( x \right )}
=\frac{1}{\frac{x^2+a x+m^2}{b x^2+a x+n^2}}
=\frac{bx^2+a x+n^2}{ x^2+b x+m^2}
=\frac{b\left ( x^{2}+\frac{a}{b}x+\frac{n^{2}}{b} \right )}{b\left ( \frac{x^{2}}{b} +x+\frac{m^{2}}{b}\right )}
= \frac{x^{2}+\frac{a}{b}x+\frac{n^{2}}{b}}{\frac{x^{2}}{b}+x+\frac{m^{2}}{b}}
 

Thus, by the above mentioned property of limits, the following is evident.

  • \lim _{x \rightarrow p} f(x)=\lim _{x \rightarrow p}\frac{x^2+a x+m^2}{b x^2+a x+n^2}= +\infty
  • \lim _{x \rightarrow p} \frac{1}{f(x)}=\lim _{x \rightarrow p}\frac{x^{2}+\frac{a}{b}x+\frac{n^{2}}{b}}{\frac{x^{2}}{b}+x+\frac{m^{2}}{b}}
Posted by

Nehul

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