Limiter en rationalisant Calculator

Apply the formula: $\lim_{x\to c}\left(a\right)$$=\lim_{x\to c}\left(a\right)$, where $a=\frac{\sqrt{5+x}-\sqrt{5}}{x}\frac{\sqrt{5+x}+\sqrt{5}}{\sqrt{5+x}+\sqrt{5}}$ and $c=0$

Apply the formula: $\frac{a}{b}\frac{c}{f}$$=\frac{ac}{bf}$, where $a=\sqrt{5+x}-\sqrt{5}$, $b=x$, $c=\sqrt{5+x}+\sqrt{5}$, $a/b=\frac{\sqrt{5+x}-\sqrt{5}}{x}$, $f=\sqrt{5+x}+\sqrt{5}$, $c/f=\frac{\sqrt{5+x}+\sqrt{5}}{\sqrt{5+x}+\sqrt{5}}$ and $a/bc/f=\frac{\sqrt{5+x}-\sqrt{5}}{x}\frac{\sqrt{5+x}+\sqrt{5}}{\sqrt{5+x}+\sqrt{5}}$

The first term ($a$) is $\sqrt{5+x}$.

The second term ($b$) is $\sqrt{5}$.

Apply the formula: $\left(a+b\right)\left(a+c\right)$$=a^2-b^2$, where $a=\sqrt{5+x}$, $b=\sqrt{5}$, $c=-\sqrt{5}$, $a+c=\sqrt{5+x}+\sqrt{5}$ and $a+b=\sqrt{5+x}-\sqrt{5}$

Apply the formula: $\left(x^a\right)^b$$=x$, where $a=\frac{1}{2}$, $b=2$, $x^a^b=\left(\sqrt{5+x}\right)^2$, $x=5+x$ and $x^a=\sqrt{5+x}$

Apply the formula: $\left(x^a\right)^b$$=x$, where $a=\frac{1}{2}$, $b=2$, $x^a^b=\left(\sqrt{5}\right)^2$, $x=5$ and $x^a=\sqrt{5}$

Evaluate the limit $\lim_{x\to0}\left(\frac{1}{\sqrt{5+x}+\sqrt{5}}\right)$ by replacing all occurrences of $x$ by $0$

Apply the formula: $a+b$$=a+b$, where $a=5$, $b=0$ and $a+b=5+0$

Combining like terms $\sqrt{5}$ and $\sqrt{5}$