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

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

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

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



12th grade math

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