Finding the limit

**Craka** · August 6th 2008, 04:26 PM

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

$ \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?

**Serena's Girl** · August 6th 2008, 05:18 PM

$ \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

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

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

$ \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:

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

**TwistedOne151** · August 6th 2008, 05:38 PM

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

--Kevin C.

**fanfan1609** · August 6th 2008, 06:02 PM

Originally Posted by TwistedOne151

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

--Kevin C.

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

**Craka** · August 6th 2008, 06:07 PM

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

**Soroban** · August 6th 2008, 09:26 PM

Hello, Craka!

Here's another approach . . .

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

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

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

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

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

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