# Math Help - Toough Limit Problem

1. ## Toough Limit Problem

Hey guys... and gals.. Got one for ya...
Find (in terms of the constant a)

2. Originally Posted by desizon
Hey guys... and gals.. Got one for ya...
Find (in terms of the constant a)
$\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h}$
$=\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}$

It should be easy now.

3. Originally Posted by General
$\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h}$
$=\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}$

It should be easy now.
I know up to that part.. This one is still causing problems.

4. Originally Posted by desizon
I know up to that part.. This one is still causing problems.
Surely, you know that If you want to multiply a fraction by another fraction, you will use this formula:
$(\frac{a}{b})(\frac{c}{d})=\frac{ac}{bd}$, for $b,d \neq 0$

Now, To multiply the numerators in your problem, Use this formula:
$(a-b)(a+b)=a^2-b^2$ .

5. I know that, So I'll have 7(a+h) - 7a on the top, which simplifies down to just

7h, since the "a" will cancel out.....

What do I do with the deal on the bottom?

6. Originally Posted by desizon
I know that, So I'll have 7(a+h) - 7a on the top, which simplifies down to just

7h, since the "a" will cancel out.....

What do I do with the deal on the bottom?
the bottom is $h(\sqrt{7(a+h)} + \sqrt{7a})$
Cancel out the h's.
and then compute your limit by direct substitution.

7. Originally Posted by desizon
Hey guys... and gals.. Got one for ya...
Find (in terms of the constant a)
Think derivatives.

8. What do I do with the deal on the bottom?
As General said, you are left with :

$\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}$

This might look impressive at first, but take the top part of the fraction :

$(\sqrt{7(a+h)} - \sqrt{7a} )(\sqrt{7(a+h)} + \sqrt{7a})$.

This can be simplified to $7(a+h) - 7a = 7(a + h - a) = 7h$. So your limit basically boils down to :

$\lim_{h\to0} \frac{7h}{h(\sqrt{7(a+h)} + \sqrt{7a})}$

Cancel the $h$, and you are pretty much done.

9. Originally Posted by desizon
Hey guys... and gals.. Got one for ya...
Find (in terms of the constant a)
It has previously been remarked to think derivatives. I will provide a further prompt: Think of differentiating $f(x) = \sqrt{7x}$ from first principles ....

10. Originally Posted by Bacterius
As General said, you are left with :

$\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}$

This might look impressive at first, but take the top part of the fraction :

$(\sqrt{7(a+h)} - \sqrt{7a} )(\sqrt{7(a+h)} + \sqrt{7a})$.

This can be simplified to $7(a+h) - 7a = 7(a + h - a) = 7h$. So your limit basically boils down to :

$\lim_{h\to0} \frac{7h}{h(\sqrt{7(a+h)} + \sqrt{7a})}$

Cancel the $h$, and you are pretty much done.
Thanks... I'm still not getting the right answer, so, what I have, is 6h left on top, and the bottom is?

11. Originally Posted by desizon
Thanks... I'm still not getting the right answer, so, what I have, is 6h left on top, and the bottom is?
how do you get 6h?

$\frac{7h}{h\left[\sqrt{7(a+h)}+\sqrt{7a}\right]} = \frac{7}{\sqrt{7(a+h)}+\sqrt{7a}}
$

now let $h \to 0$

12. edit:::
Thanks all. Got it.