# How to solve $\lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}$?

How to solve the following question?

\begin{eqnarray} \\\lim_{x\to 0}f(x)&=&\lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}\\ \\&=&\lim_{x\to 0}\frac{\sqrt{\frac{x^2+4x+5}{x^2}}-\sqrt{\frac{5}{x^2}}}{1}\\ \\&=&\frac{\sqrt{\lim_{x\to 0}(1+\frac{4}{x}+\frac{5}{x^2})}-\sqrt{\lim_{x\to 0}(\frac{5}{x^2})}}{1} \\&=& \infty-\infty \\&=&? \end{eqnarray}

It can be taken limit and $\lim_{x\to 0}f(x)=0.89...$ but when x=0, y do not exist. Rgiht?

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HINT: Conjugate when $x \to 0$. –  Don Larynx Oct 18 '13 at 11:42

$$\lim_{x\to 0}f(x) = \lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}$$

Evaluating the limit of $f(x)$ as $x\to 0$ is not be be confused with evaluating $f(x)$ at zero, where, as you may have noted in this case, $f(x)$ is not defined. What we are interested in when taking the limit as $x\to 0$ is the behavior of $f(x)$ as $x$ approaches, gets arbitrarily close to, zero. If it converges to some value, then that is the limit. This is not to be mistaken with $f(0)$, even though many times, you'll find functions for which $\lim_{x \to 0} f(x) = f(0).$

In this case, since we have an indeterminate form $\frac 00$, we can use L'hopital, or we can multiply the numerator and denominator by the conjugate of the numerator to obtain a non-indeterminate form:

\begin{align} \\\lim_{x\to 0}f(x) & =\lim_{x\to 0}\frac{\sqrt{x^2+4x+5}-\sqrt{5}}{x}\\ \\& =\lim_{x\to 0}\frac{\sqrt{x^2+4x+5} - \sqrt 5}{x}\cdot \frac{\sqrt{x^2 + 4x + 5} + \sqrt 5}{\sqrt{x^2 + 4x + 5} + \sqrt 5}\\ \\ &= \lim_{x\to 0} \frac{(x^2 + 4x + 5) - 5}{x(\sqrt{x^2 + 4x + 5} + \sqrt 5)}\\ \\ & = \lim_{x\to 0} \frac{x(x + 4)}{x(\sqrt{x^2 + 4x + 5} + \sqrt 5)}\\ \\ & = \lim_{x\to 0} \frac{x + 4}{\sqrt{x^2 + 4x + 5} + \sqrt 5} \\ \\ & = \frac 4{2\sqrt 5} = \frac 2{\sqrt 5} \end{align}

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Rationalize the Numerator

$$\sqrt{x^2+4x+5}-\sqrt5=\frac{x^2+4x+5-5}{\sqrt{x^2+4x+5}+\sqrt5}=\frac{x(x+4)}{\sqrt{x^2+4x+5}+\sqrt5}$$

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