What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$?

Question

I tried multiplying by the conjugate which gave:

$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$

But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there?

Answer 1

$\require{cancel}$

Multiply by the conjugate of $\sqrt{x+2}-2$ which is $\sqrt{x+2}+2$, was a good call; you just made an error in your multiplication by the conjugate.

After multiplying top and bottom by $\sqrt{x +2} + 2$ we get $$\frac{x-2}{(x-2)(\sqrt{x+2}+2)}$$

Cancel the factor $(x-2)$ which appears in the numerator and in the denominator, $$\frac{\cancel{x-2}}{(\cancel{x-2})(\sqrt{x+2}+2)}$$to get $$\lim_{x\to 2} \frac 1{\sqrt{x+2}+2}\; = \;\frac 14$$

Answer 2

\begin{align}\lim_{x \to 2}\frac{\sqrt{x+2}-2}{x-2}&= \lim_{x \to 2}\frac{\sqrt{x+2}-2}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)} \\ &=\lim_{x \to 2}\frac{1}{\sqrt{x+2}+2}\end{align}

Answer 3

Let $x=2+h$ and let $h \rightarrow 0$ hence $$ \frac{\sqrt{4+h}-2}{h}=\frac{2\sqrt{1+\frac{h}{4}}-2}{h}=\frac{2\left(1+\frac{h}{8}-1+o\left(h\right)\right)}{h} \underset{h \rightarrow 0}{\rightarrow}\frac{1}{4} $$ Hence

$$ \frac{\sqrt{x+2}-2}{x-2} \underset{x \rightarrow 2}{\rightarrow}\frac{1}{4}] $$

Answer 4

Hint $$\frac{\sqrt{x+2}-2}{x-2}\cdot \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}= \frac{\sqrt{x+2}^2-2^2}{(x-2)\sqrt{x+2}+2}$$ $$ \frac{x-2}{(x-2)\sqrt{x+2}+2}= \frac{1}{\sqrt{x+2}+2}$$

Answer 5

It's the rate of variation at $x=2$ of the function $\sqrt{x+2}$, hence its limit is the derivative: $$(\sqrt{x+2})'_{x=2}=\frac1{2\sqrt{x+2}}\biggr|_{x=2} = \frac14.$$

Answer 6

Note that the limit is by definition the derivative of

$$f(x)=\sqrt{x+2}\implies f'(x)=\frac1{2\sqrt{x+2}}$$

thus

$$\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}=\frac{1}{2\sqrt{2+2}}=\frac14$$

Answer 7

For fun:

$\dfrac{\sqrt{x+2} -2}{(\sqrt{x+2})^2-4}=$

$\dfrac{\sqrt{x+2}-2}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}=$

$\dfrac{1}{\sqrt{x+2}+2}.$

The limit $x \rightarrow 2$ is?