In mathematics, $f(x)$ is a function and $\dfrac{1}{f(x)}$ is its reciprocal, where $x$ represents a variable. Assume, $a$ is a constant and represents a value of $x$. The limits of both functions can be written in mathematical form as $x$ approaches $a$.

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

$(2)\,\,\,$ $\displaystyle \large \lim_{x \,\to\, a}{\normalsize \dfrac{1}{f(x)}}$

Now, let’s prove the reciprocal rule of limits in calculus mathematically by using them.

Now, evaluate the limit of the function $f(x)$ as $x$ approaches $a$ by the direct substitution.

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

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

Now, evaluate the limit of the multiplicative inverse function $f(x)$ as $x$ approaches $a$ by the direct substitution method.

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

The limit rule for a reciprocal of a function can be derived by considering the equations of the above two steps. We have derived an equation in the above step as follows.

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

In first step, we have also derived that

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

Therefore, we can replace the value of function $f(a)$ by its equivalent expression in the above equation.

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

Therefore, it is proved that the limit of a reciprocal of a function is equal to the reciprocal of the limit of the function.

Latest Math Topics

Jul 24, 2022

Jul 15, 2022

Latest Math Problems

Jul 29, 2022

Jul 17, 2022

Jun 02, 2022

Apr 06, 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