The process of evaluating the limit of a function by rationalizing the expression in irrational form is called the method of evaluating the limits by rationalization.

The irrational form expressions are involved in forming the functions in some cases in mathematics. Sometimes, the limits of such functions are indeterminate mainly due to the involvement of the expressions with radical symbols.

The limit as the indeterminate can be avoided by rationalizing the irrational form expression in the function, and it is called the method of evaluating the limits by rationalisation.

There are three simple steps to find the limit of a function by rationalization.

- Search for the irrational expressions in the function.
- Evaluate the limit by direct substitution to know whether the radical form expressions play a role in giving the indeterminate form.
- Rationalize the irrational expression by its conjugate, and then evaluate it by the direct substitution.

Let’s understand the process of evaluating the limits by rationalization.

$\displaystyle \large \lim_{x\,\to\,6}{\normalsize \dfrac{\sqrt{x}-\sqrt{6}}{x-6}}$ $\,=\,$ $\dfrac{0}{0}$

According to the direct substitution, the limit a rational function is indeterminate. It is mainly due to the involvement of an expression in radical form in the numerator. Let’s observe what happens after rationalizing the numerator by its conjugate function.

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,6}{\normalsize \Bigg(\dfrac{\sqrt{x}-\sqrt{6}}{x-6} \times 1\Bigg)}$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,6}{\normalsize \Bigg(\dfrac{\sqrt{x}-\sqrt{6}}{x-6} \times \dfrac{\sqrt{x}+\sqrt{6}}{\sqrt{x}+\sqrt{6}}\Bigg)}$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,6}{\normalsize \dfrac{\big(\sqrt{x}-\sqrt{6}\big) \times \big(\sqrt{x}+\sqrt{6}\big)}{(x-6) \times \big(\sqrt{x}+\sqrt{6}\big)}}$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,6}{\normalsize \dfrac{\big(\sqrt{x}\big)^2-\big(\sqrt{6}\big)^2}{(x-6) \times \big(\sqrt{x}+\sqrt{6}\big)}}$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,6}{\normalsize \dfrac{x-6}{(x-6) \times \big(\sqrt{x}+\sqrt{6}\big)}}$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,6}{\normalsize \dfrac{\cancel{x-6}}{\cancel{(x-6)} \times \big(\sqrt{x}+\sqrt{6}\big)}}$

$=\,\,$ $\displaystyle \large \lim_{x\,\to\,6}{\normalsize \dfrac{1}{\sqrt{x}+\sqrt{6}}}$

$=\,\,$ $\dfrac{1}{\sqrt{6}+\sqrt{6}}$

$=\,\,$ $\dfrac{1}{2\sqrt{6}}$

Initially, the limit of a function is indeterminate but the rationalization method changed the result. Hence, the limit of a function can be evaluated by the rationalization when the irrational form expressions are involved in the functions.

$(1).\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt{1+x}-\sqrt{1-x}}{2x}}$

$(2).\,\,$ $\displaystyle \large \lim_{x\,\to\,-1}{\normalsize \dfrac{\sqrt[\Large 3]{7-x}-2}{x+1}}$

$(3).\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Big(\sqrt{x^2+2x}-x\Big)}$

The list of questions on limits by rationalization for practice with solutions to learn how to find the limit of a function by rationalisation.

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved