Reciprocal rule of Limits

Math Doubts

Formula

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

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

Introduction

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.

Example

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

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