Toough Limit Problem

**desizon** · January 17th 2010, 11:12 AM

Hey guys... and gals.. Got one for ya...

Find (in terms of the constant a)

**General** · January 17th 2010, 11:17 AM

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.

**desizon** · January 17th 2010, 02:24 PM

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.

**General** · January 17th 2010, 02:28 PM

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:
$(a-b)(a+b)=a^2-b^2$ .

**desizon** · January 17th 2010, 05:51 PM

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?

**General** · January 17th 2010, 06:23 PM

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.

**Drexel28** · January 17th 2010, 07:57 PM

Originally Posted by desizon

Hey guys... and gals.. Got one for ya...

Find (in terms of the constant a)

Think derivatives.

**Bacterius** · January 17th 2010, 08:12 PM

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.

**mr fantastic** · January 18th 2010, 12:52 AM

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 ....

**desizon** · January 18th 2010, 08:05 AM

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?

**skeeter** · January 18th 2010, 08:17 AM

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?

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

now let $h \to 0$

**desizon** · January 18th 2010, 12:15 PM

edit:::
Thanks all. Got it.

Thread: Toough Limit Problem

LinkBack

Thread Tools

Display

Toough Limit Problem

Similar Math Help Forum Discussions

I have a problem in understanding a problem about the limit of a function.

limit question on a limit comparison test problem.

Limit, Limit Superior, and Limit Inferior of a function

Limit Problem

limit problem

Search Tags