# Thread: Finding the limit

1. ## Finding the limit

Question is find the value of
[tex]
\mathop {\lim }\limits_{x \to 9} \frac{{t - 9}}{{3 - \sqrt t }}
/MATH]

This is as far as I get

$\displaystyle \mathop {\lim }\limits_{x \to 9} \frac{{t - 9}}{{3 - \sqrt t }} \\ = \mathop {\lim }\limits_{x \to 9} \frac{{t - 9}}{{3 - \sqrt t }} \times \frac{{3 + \sqrt t }}{{3 + \sqrt t }} \\ = \mathop {\lim }\limits_{x \to 9} \frac{{t^{\frac{3}{2}} + 3t - 9\sqrt t - 27}}{{9 - t}} \\$

At this stage I'm left with a divide by zero case, if I sub in 9 in an attempt to find the limit. Where do I go from here?

2. $\displaystyle \mathop {\lim }\limits_{t \to 9} \frac{{t - 9}}{{3 - \sqrt t }}$

Since you end up with a 0/0 indeterminate case, you can use L'Hospital's rule to solve the limit. That is, if you are given

$\displaystyle \mathop {\lim }\limits_{x \to a} \frac{f(x)}{g(x)}$

and you end up with 0/0, the limit is

$\displaystyle \mathop {\lim }\limits_{x \to a} \frac{f'(x)}{g'(x)}$

So take the derivative of the numerator, then the derivative of the denominator, then solve for the limit:

$\displaystyle \mathop {\lim }\limits_{t \to 9} \frac{1}{- 0.5t^{-0.5}} = \mathop {\lim }\limits_{t \to 9} -2 \sqrt {t} = -6$

3. You could also not multiply out the numerator:
$\displaystyle \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}\cdot\frac{3+\sqrt{t}}{3+\sqrt{t}}=\lim_{ x \to 9} \frac{(t-9)(3+\sqrt{t})}{9 - t}$
And since t-9 and 9-t are opposites, their quotient is -1, and we get that the above
$\displaystyle \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x \to 9}-(3+\sqrt{t})=-6$.

--Kevin C.

4. Originally Posted by TwistedOne151
You could also not multiply out the numerator:
$\displaystyle \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}\cdot\frac{3+\sqrt{t}}{3+\sqrt{t}}=\lim_{ x \to 9} \frac{(t-9)(3+\sqrt{t})}{9 - t}$
And since t-9 and 9-t are opposites, their quotient is -1, and we get that the above
$\displaystyle \lim_{x\to{9}}\frac{t-9}{3-\sqrt{t}}=\lim_{x \to 9}-(3+\sqrt{t})=-6$.

--Kevin C.
I think if we note that $\displaystyle (t-9)=-(3+\sqrt{t})(3-\sqrt{t})$ then it is easier.

5. Thanks Kevin. I've seen that done before but I couldn't remember what to do thanks.

6. Hello, Craka!

Here's another approach . . .

$\displaystyle \lim_{t\to9}\, \frac{t - 9}{3 - \sqrt{t}}$

Factor -1 out of the denominator: .$\displaystyle \lim_{t\to9}\,\frac{t-9}{-(\sqrt{t} - 3)}$

Factor the numerator: .$\displaystyle \lim_{t\to9}\,-\frac{(\sqrt{t} - 3)(\sqrt{t} + 3)}{\sqrt{t}-3}$

Reduce: .$\displaystyle \lim_{t\to9}\,-(\sqrt{t} + 3)$

Evaluate: .$\displaystyle -(\sqrt{9} + 3) \;=\;-6$