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  • If  <img alt="\lim _{x \rightarrow p} f(x)=+\infty \text { where } f(x)=\frac{x^3+a b x+m^2}{b y^3+a y+n^2}, b \neq 0 \quad \text {, then } \lim _{y \rightarrow p} \frac{y^3+\frac{a}{b} y+\fra

If  \lim _{x \rightarrow p} f(x)=+\infty \text { where } f(x)=\frac{x^3+a b x+m^2}{b y^3+a y+n^2}, b \neq 0 \quad \text {, then } \lim _{y \rightarrow p} \frac{y^3+\frac{a}{b} y+\frac{n^2}{b}}{\frac{x^3}{b}+a x+\frac{m^2}{b}}=\text { ? }

Option: 1

+\infty


Option: 2

-\infty


Option: 3

0


Option: 4

Cannot be determined


Answers (1)

best_answer

The following property of limits is important.
When the limit is defined as \lim_{y\rightarrow 1}f\left ( y \right )=+\infty, then the limit \lim_{y\rightarrow 1}\frac{1}{f\left ( y \right )}=0.

The provided function is f\left ( x \right )= \frac{x^{3}+abx+m^{2}}{by^{3}+ay+n^{2}}.

So, the function \frac{1}{f\left ( x \right )}  is as follows:

\frac{1}{f\left ( x \right )}
= \frac{1}{\frac{x^{3}+abx+m^{2}}{by^{3}+ay+n^{2}}}
= \frac{by^{3}+ay+n^{2}}{x^{3}+abx+m^{2}}
= \frac{b\left ( y^{3}+\frac{a}{b}y+\frac{n^{2}}{b} \right )}{b\left ( \frac{x^{3}}{b}+ax+\frac{m^{2}}{b} \right )}
=\frac{ y^{3}+\frac{a}{b}y+\frac{n^{2}}{b} }{\frac{x^{3}}{b}+ax+\frac{m^{2}}{b}}
Thus, by the above mentioned property of limits, the following is evident.

  • \lim_{x\rightarrow p}f\left ( x \right )= \lim_{x\rightarrow p}\frac{x^{2}+ax+m^{2}}{bx^{2}+ax+n^{2}}= +\infty
  • \lim_{x\rightarrow p}\frac{1}{f\left ( x \right )}= \lim_{x\rightarrow p}\frac{y^{3}+\frac{a}{b}y+\frac{n^{2}}{b}}{\frac{x^{3}}{b}+ax+\frac{m^{2}}{b}}= 0


It is important to note here that the provided limit in the expression \lim_{x\rightarrow p}\frac{y^{3}+\frac{a}{b}y+\frac{n^{2}}{b}}{\frac{x^{3}}{b}+ax+\frac{m^{2}}{b}} is y\rightarrow p  are not x\rightarrow p, and hence it cannot be determined

Posted by

Ritika Kankaria

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