3
$\begingroup$

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 $3$.

$\endgroup$
7
$\begingroup$

The solution goes as follows:

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

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Peter Mortensen Jan 17 '16 at 16:51
8
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $$\lim_{y\to0}\dfrac1{y(y+6)^2+6(y+6)}=?$$

$\endgroup$
1
$\begingroup$

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 $$ \lim_{x\to 3}\frac{\sqrt{x+6}-3}{x-3}= \lim_{x\to3}\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x-3)(\sqrt{x+6}+3)} =\lim_{x\to3}\frac{x-3}{(x-3)(\sqrt{x+6}+3)}=\dots $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.