Finding Limit by Rationalizing

In this section we will discuss finding limit by rationalizing technique. This method involves rationalizing the numerator or denominator. Rationalizing the numerator or denominator means the multiplying numerator or denominator by the conjugate of numerator or denominator.
For example, the conjugate of $\sqrt{x}+4$ is $\sqrt{x} - 4$ ; the conjugate of $\sqrt{x + 2 } - 3$ is $\sqrt{x + 2 } + 3 $

Examples on finding limit by rationalizing method

1) Find the limit: $\lim_{x->0}\frac{\sqrt{x+1}-1}{x}$

Solution : $\lim_{x->0}\frac{\sqrt{x+1}-1}{x}$ , if we plug in direct x = 0 so we will get $\frac{0}{0}$, so direct substitution fails.

Here, we will will rationalizing technique.As there is a radical expression in the numerator so we will rationalize the numerator. We will multiply numerator and denominator by ${\sqrt{x+1}+1}$

$\lim_{x->0}\frac{\sqrt{x+1}-1}{x}$ = $\lim_{x->0}\left ( \frac{\sqrt{x+1}-1}{x} \right )\left ( \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} \right )$

= $\lim_{x->0}\left ( \frac{{x+1}-1}{x(\sqrt{x+1}+1)} \right )$

= $\lim_{x->0}\left ( \frac{x}{x(\sqrt{x+1}+1)} \right )$

$\lim_{x->0}\frac{\sqrt{x+1}-1}{x}$ = $\lim_{x->0}\left ( \frac{1}{\sqrt{x+1}+1} \right )$

Now plug in x = 0
$\lim_{x->0}\frac{\sqrt{x+1}-1}{x}$ = $\left ( \frac{1}{\sqrt{0+1}+1} \right )$

$\lim_{x->0}\frac{\sqrt{x+1}-1}{x}$ = $\left ( \frac{1}{1 + 1} \right )$

$\lim_{x->0}\frac{\sqrt{x+1}-1}{x}$ = $\left ( \frac{1}{2} \right )$

2) Find the limit: $\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$

Solution : $\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$ , if we plug in direct x = 0 so we will get $\frac{0}{0}$, so direct substitution fails.

Here, we will will rationalizing technique.As there is a radical expression in the numerator so we will rationalize the numerator. We will multiply numerator and denominator by ${\sqrt{2+x}+\sqrt2}$

$\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$ = $\lim_{x->0}\left (\frac{\sqrt{2+x}-\sqrt2}{x} \right )\left ( \frac{\sqrt{2+x}+\sqrt2}{\sqrt{2+x}+\sqrt2} \right )$

= $\lim_{x->0}\left (\frac{2+x-2}{x(\sqrt{2+x}+\sqrt2} \right )$

$\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$ = $\lim_{x->0}\left (\frac{x}{x(\sqrt{2+x}+\sqrt2} \right )$

$\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$ = $\lim_{x->0}\left (\frac{1}{(\sqrt{2+x}+\sqrt2} \right )$

Now plug in x = 0

$\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$ = $\left (\frac{1}{(\sqrt{2+0}+\sqrt2} \right )$

$\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$ = $\left (\frac{1}{\sqrt{2}+\sqrt2} \right )$

$\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$ = $\left (\frac{1}{(2\sqrt{2}}\right )$

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

Russia-Ukraine crisis update - 3rd Mar 2022

The UN General assembly voted at an emergency session to demand an immediate halt to Moscow's attack on Ukraine and withdrawal of Russian troops.



12th grade math

From finding limit by rationalizing to Home