# Finding limit/derivative

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• Mar 25th 2008, 03:22 PM
theowne
Finding limit/derivative
I have two questions, both are kind of similar involving rationalization and I would appreciate if someone can push me in the correct direction.

The first is finding the limit of this:
sqrt(3x-5) - 4
x-7

And this is as limit x-> 7

I tried to rationalize and got this in the end

3(x-7)
(x-7)(sqrt(3x-5) - 4)

But after canceling out the (x-7), the denominator is still 0. I'm not sure where I can go from there, I think I must have made a mistake somewhere..

_______________________________________

The second question is specifically finding the slope to the tangent at a given point from first principles.

of y = sqrt{2x+8} at (-2,2)

So this was the form I used,

Limit h->0

sqrt{2(2+h)+8} - 2
h

And then after trying to rationalize I get this

2(2+h) + 8 - 4
h(sqrt(2(2+h) +8) +2)

I don't see where I can go from here either. The top doesn't leave something in kh form so I can't cancel it out....Did I make a mistake or approach it from the wrong direction?
• Mar 25th 2008, 03:37 PM
mr fantastic
Quote:

Originally Posted by theowne
I have two questions, both are kind of similar involving rationalization and I would appreciate if someone can push me in the correct direction.

The first is finding the limit of this:
sqrt(3x-5) - 4
x-7

And this is as limit x-> 7

I tried to rationalize and got this in the end

3(x-7)
(x-7)(sqrt(3x-5) - 4) Mr F says: That - should be a +.

But after canceling out the (x-7), the denominator is still 0. I'm not sure where I can go from there, I think I must have made a mistake somewhere..

_______________________________________

The second question is specifically finding the slope to the tangent at a given point from first principles.

of y = sqrt{2x+8} at (-2,2)

So this was the form I used,

Limit h->0

sqrt{2(2+h)+8} - 2
h

Mr F says: x = -2 so the bit in red should be -2. So you need to re-work the solution from here.

[snip]

..
• Mar 25th 2008, 03:48 PM
theowne
I don't understand why it should be -(-2). The form is f(x+h) - f(x) correct (over h)? The components say (-2,2) so isn't f(x)=2?
• Mar 25th 2008, 03:53 PM
mr fantastic
Quote:

Originally Posted by theowne
I don't understand why it should be -(-2). The form is f(x+h) - f(x) correct (over h)? The components say (-2,2) so isn't f(x)=2?

I made a typo. I've edited my reply. Take another look now.

x = -2 so you need f(-2 + h) - f(-2). You have f(2 + h) - f(-2).
• Mar 25th 2008, 03:55 PM
theowne
Thanks for the help. I seem to make these kind of simple errors often.
• Mar 25th 2008, 05:27 PM
theowne
If I may ask one more question, I have one more example:

2x^3 - 4x
3x^4 + 2x^3

I'm not too sure how to approach this question. I'm guessing it involves factoring but I'm not sure exactly what..
• Mar 25th 2008, 05:37 PM
topsquark
Quote:

Originally Posted by theowne
If I may ask one more question, I have one more example:

2x^3 - 4x
3x^4 + 2x^3

I'm not too sure how to approach this question. I'm guessing it involves factoring but I'm not sure exactly what..

(Angry) New questions belong in new threads!

I presume you need to simplify this?
$\displaystyle \frac{2x^3 - 4x}{3x^4 + 2x^3}$

$\displaystyle = \frac{2x(x^2 - 4)}{x^3(3x + 2)} = \frac{2(x^2 - 4)}{x^2(3x + 2)}$

Note that the numerator can be factored further, but as nothing more will cancel I didn't worry about doing it.

-Dan
• Mar 25th 2008, 05:40 PM
theowne
Hi,

Need to find the limit as x approaches 0. Sorry, I didn't realize I'd need a new thread since this question was similar to the originals. Simplified that way, the denominator would still be zero, wouldn't it?
• Mar 26th 2008, 04:33 AM
topsquark
Quote:

Originally Posted by theowne
Hi,

Need to find the limit as x approaches 0. Sorry, I didn't realize I'd need a new thread since this question was similar to the originals. Simplified that way, the denominator would still be zero, wouldn't it?

Yes, and as the limit in the numerator does not go to zero, then the limit does not exist.

-Dan
• Mar 26th 2008, 04:39 AM
mr fantastic
Quote:

Originally Posted by topsquark
(Angry) New questions belong in new threads!

I presume you need to simplify this?

[snip]
-Dan

And ideally you will post the whole question so that presumptions/assumptions are unnecessary.