$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{1}{f(x)}}$ $\,=\,$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, a}{\normalsize f(x)}}$

The limit of a reciprocal function equals to the reciprocal of its limit is called the reciprocal rule of limits.

Let $f(x)$ be a function in terms of $x$ and its multiplicative inverse is written as $\dfrac{1}{f(x)}$. When $x$ approaches a value $a$, the limit of the reciprocal function is written as follows.

$\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{1}{f(x)}}$

Mathematically, the limit of the reciprocal of a function is equal to the reciprocal of the limit of function as $x$ tends to $a$.

$\implies$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{1}{f(x)}}$ $\,=\,$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, a}{\normalsize f(x)}}$

The property of the equality is called the reciprocal rule of limits.

Evaluate $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{1}{1+x^2}}$

Now, evaluate the rational function as $x$ approaches $2$ by the direct substitution method.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{1}{1+x^2}}$ $\,=\,$ $\dfrac{1}{1+(2)^2}$ $\,=\,$ $\dfrac{1}{5}$

It is evaluated that the limit of the rational expression as $x$ approaches $2$ is equal to $\dfrac{1}{5}$.

Now, let us evaluate reciprocal of limit of the function $1+x^2$ as $x$ approaches $2$ by the direct substitution.

$\implies$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 2}{\normalsize 1+x^2}}$ $\,=\,$ $\dfrac{1}{1+(2)^2}$ $\,=\,$ $\dfrac{1}{5}$

It is evaluated that the multiplicative inverse of the limit of the function $1+x^2$ as $x$ approaches $2$ is also equal to $\dfrac{1}{5}$.

Therefore, we can understood that

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{1}{1+x^2}}$ $\,=\,$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 2}{\normalsize 1+x^2}}$ $\,=\,$ $\dfrac{1}{5}$

Learn how to prove the reciprocal property of limits in calculus four using it as a formula in mathematics.

Latest Math Topics

Jan 06, 2023

Jan 03, 2023

Jan 01, 2023

Dec 26, 2022

Dec 08, 2022

Latest Math Problems

Jan 31, 2023

Nov 25, 2022

Nov 02, 2022

Oct 26, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved