Limits of functions with square roots in the denominator For the function below, show that the $\displaystyle\lim_{x \to -2} f(x) = 4$, and justify your answer. (without using L'Hôpital's rule).
$$f(x)= \dfrac{x+2}{\sqrt{6+x}-2}$$ 
My attempt is as follows:
Since $f(x)$ is defined when $6+x>0$, i.e so long as $x>-6$, the function is defined in the neighborhood of $-2$ and the limit does indeed exist and so we can proceed... 
(i don't know what the method is to prove this limit when we have a square root). What is the approach you would go about to show this?     
 A: An usual way to deal with square root is changing of variable: since the square root looks annoying, we may set it as a new variable $t$,thus eliminating square root ---- this principle might also be helpful in integration problems. For this problem, if set $t = \sqrt{x + 6}$, then $x = t^2 - 6$, and as $x \to -2$, $t \to 2$. Thus
\begin{align}
& \lim_{x \to -2}\frac{x + 2}{\sqrt{x + 6} - 2} \\
= & \lim_{t \to 2} \frac{t^2 - 6 + 2}{t - 2} \\
= & \lim_{t \to 2} \frac{(t + 2)(t - 2)}{t - 2} \\
= & \lim_{t \to 2} t + 2 \\
= & 4.
\end{align}
A: Hint One option is to recognize $\frac{1}{f(x)}$ as the difference quotient for a particular function at a particular point and use the definition of derivative.
A second option is to multiply both the numerator and denominator by the conjugate of the denominator, namely, $\sqrt{6 + x} + 2$, and then simplify.
A: It is sure that multiplying by the conjugate of the denominator makes the problem simple when only the limit is required.
Just for your curiosity, let me show you another method will would allow to solve the problem in a quite simple manner.
First, change $x=y-2$  $$f= \dfrac{x+2}{\sqrt{6+x}-2}=\dfrac{y}{\sqrt{4+y}-2}=\frac12\dfrac{y}{\sqrt{1+\frac y4}-1}$$ Now, consider the Taylor series $$\sqrt{1+z}=1+\frac{z}{2}-\frac{z^2}{8}+O\left(z^3\right)$$ and replace $z$ by $\frac y4$. This gives $$\sqrt{1+\frac y4}=1+\frac{y}{8}-\frac{y^2}{128}+O\left(y^3\right)$$ which makes $$f=\frac 12\frac y{1+\frac{y}{8}-\frac{y^2}{128}+O\left(y^3\right)-1}=\frac 12\frac 1 {\frac{1}{8}-\frac{y}{128}+O\left(y^2\right)}$$ and now, using long division $$f=4+\frac{y}{4}+O\left(y^2\right)=4+\frac{x+2}{4}+O\left((x+2)^2\right)$$ which shows the limit and how it is approached.
A: Hint: You probably know that $(a+b)(a-b)=a^2-b^2$. This is particularly useful if $a$ or $b$ (or both) is a square root, since the squaring of each term will make the radical "go away".
So if you have a difference of root(s) in the denominator, you can supply the other factor (sum of the same root(s)) in both numerator and denominator to achieve an effective squaring  of the terms in the denominator.
The trade-off is that you now have roots in the numerator, but that's often easier to deal with.
I hope this helps.
