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?
Best Answer
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}