For example, the conjugate of $\sqrt{x}+4$ is $\sqrt{x} - 4$ ; the conjugate of $\sqrt{x + 2 } - 3$ is $\sqrt{x + 2 } + 3 $

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: $\lim_{x->0}\frac{\sqrt{2+x}-\sqrt2}{x}$

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

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

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

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

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