# Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.

The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.

I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.

Now how do I rationalize the numerator?

It is giving the $\frac{0}{0}$ form after plugging in $3$.

The solution goes as follows:

$$\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$$ $$=\lim_\limits{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x^2-9}\right)$$ $$=\lim_\limits{x\to 3}\left[1+\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x^2-9)(\sqrt{x+6}+3)}\right]$$ $$=\lim_\limits{x\to 3}\left[1+\frac{x+6-9}{(x^2-9)(\sqrt{x+6}+3)}\right]$$ $$=\lim_\limits{x\to 3}\left[1+\frac{x-3}{(x-3)(x+3)(\sqrt{x+6}+3)}\right]$$ $$=\lim_\limits{x\to 3}\left[1+\frac{1}{(x+3)(\sqrt{x+6}+3)}\right]$$ $$=1+\frac{1}{(3+3)(\sqrt{3+6}+3)}$$ $$=\frac{37}{36}$$

• Given that it is most likely a homework question, what is the gist of it? What would be a more general problem-solving approach? Perhaps the teacher is not interested in the result, but in the why and the reasoning. – Peter Mortensen Jan 17 '16 at 16:51

Hint. You may write, as $x \to 3$, $$\frac{\sqrt{x+6}-3}{x^2-9}=\frac{(x+6)-9}{(x-3)(x+3)(\sqrt{x+6}+3)}=\frac1{(x+3)(\sqrt{x+6}+3)}$$

Set $\sqrt{x+6}-3=y\implies x=(y+3)^2-6=y^2+6y+3$

$$\implies\lim_{x\to3}\dfrac{\sqrt{x+6}-3}{x^2-9}$$

$$=\lim_{y\to0}\dfrac y{(y^2+6y+3)^2-9}$$

$$=\lim_{y\to0}\dfrac y{(y^2+6y)^2+6(y^2+6y)}$$

As $y\to0,y\ne0$

So, cancel out $y$ from N & D to get $$\lim_{y\to0}\dfrac1{y(y+6)^2+6(y+6)}=?$$

Your first step is good. Now you want to compute $$\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}$$ because the factor $x+3$ at the denominator poses no problem. This should remind you of the definition of derivative and indeed it's the derivative of $f(x)=\sqrt{x+6}$ at $3$. Since $$f'(x)=\frac{1}{2\sqrt{x+6}}$$ for $x>-6$, you have $f'(3)=1/6$ and so $$\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}= \lim_{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x-3}\frac{1}{x+3}\right)= 1+\frac{1}{6}\frac{1}{6}$$

If you don't know about derivatives, just do $$\lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}= \lim_{x\to3}\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x-3)(\sqrt{x+6}+3)} =\lim_{x\to3}\frac{x-3}{(x-3)(\sqrt{x+6}+3)}=\dots$$