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.

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