# A limit as x approaches a.

• Feb 11th 2011, 12:09 AM
qleeq
A limit as x approaches a.
$\lim_{x \to a} \dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}$

There aren't any steps for this on Wolfram so..

I have a question. I know that this limit is to find the derivative at a certain point, whereas $x \to 0$ is for any point. I must have forgotten how to evaluate this but.. How exactly do I do this? Can I use direct substitution? I'm not looking for the answer or for you to show me how you would do it, just where I start. Thanks!
• Feb 11th 2011, 12:14 AM
Prove It
Since this goes to $\displaystyle \frac{0}{0}$ you can use L'Hospital's Rule.
• Feb 11th 2011, 12:18 AM
qleeq
Quote:

Originally Posted by Prove It
Since this goes to $\displaystyle \frac{0}{0}$ you can use L'Hospital's Rule.

Yes but since the derivatives of those constant a's are zero isn't that a problem?
• Feb 11th 2011, 12:18 AM
Chris L T521
Quote:

Originally Posted by qleeq
$\lim_{x \to a} \dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}$

There aren't any steps for this on Wolfram so..

I have a question. I know that this limit is to find the derivative at a certain point, whereas $x \to 0$ is for any point. I must have forgotten how to evaluate this but.. How exactly do I do this? Can I use direct substitution? I'm not looking for the answer or for you to show me how you would do it, just where I start. Thanks!

I don't see how this limit could represent the derivative of some function. So to evaluate this, you could try to use L'Hôpital's rule, since $\lim\limits_{x\to a}\dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}\rightarrow\dfrac{0}{0}$.

Edit: Prove It beat me to it.
• Feb 11th 2011, 12:31 AM
Prove It
They are not constant functions, they are coefficients of the functions of $\displaystyle x$, which is fine...
• Feb 11th 2011, 12:37 AM
qleeq
Quote:

Originally Posted by Chris L T521
I don't see how this limit could represent the derivative of some function. So to evaluate this, you could try to use L'Hôpital's rule, since $\lim\limits_{x\to a}\dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}\rightarrow\dfrac{0}{0}$.

Edit: Prove It beat me to it.

Sorry, this is only my first year of calculus and we haven't had a formal lecture on L'Hopital's rule yet, so I'm only just now learning how it works.
• Feb 11th 2011, 12:53 AM
Prove It
Quote:

Originally Posted by qleeq
Sorry, this is only my first year of calculus and we haven't had a formal lecture on L'Hopital's rule yet, so I'm only just now learning how it works.

That's fine...

Basically $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$ as long as your original function gives $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ from direct substitution.

I can show you a proof of this if you like.
• Feb 11th 2011, 01:09 AM
qleeq
I have a feeling I'm over-complicating this..

$\lim_{x \to a} \dfrac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{ax^3}}$

$\lim_{x \to a} \dfrac{\dfrac{1}{2}(2a^3x-x^4)^{-1/2} - \dfrac{1}{3}a(a^2x)^{-2/3}}{a - \frac{1}{4}(ax^3)^{-3/4}}$

$\lim_{x \to a} \dfrac{\dfrac{1}{2}(2a^3x-x^4)^{-1/2}\*(2a^3 - 4x^3) - \dfrac{1}{3}a(a^2x)^{-2/3}\*a^2}{a - \dfrac{1}{4}(ax^3)^{-3/4}\*(3x^2a)}$
• Feb 11th 2011, 01:11 AM
Prove It
You haven't done the differentiating properly... You need to use the Chain Rule for each term...
• Feb 11th 2011, 01:14 AM
qleeq
Quote:

Originally Posted by Prove It
You haven't done the differentiating properly... You need to use the Chain Rule for each term...

Lame. Where exactly have I messed up?
• Feb 11th 2011, 01:16 AM
Prove It
Each term you've differentiated is incorrect...
• Feb 11th 2011, 02:14 AM
qleeq
Figured just as much. I think the thing that's throwing me off is the 'a'.. how do I deal with it?
• Feb 11th 2011, 02:55 AM
Prove It
$\displaystyle a$ gets treated as a constant.
• Feb 11th 2011, 04:48 AM
HallsofIvy
No, it is NOT the "a" that is throwing you off. What is throwing you haven't used the chain rule as Prove It said.

You have the derivative of $\sqrt{2a^3x- x^4}= \left(2a^3x- x^4\right)^{1/2}$ as $\frac{1}{2}\left(2a^3x- x^5\right)^{-1/2}$. That's wrong because, by the chain rule, you must multiply that by the derivative of $2a^x- x^4$ with respect to x.
• Feb 11th 2011, 05:03 AM
qleeq
Quote:

Originally Posted by HallsofIvy
No, it is NOT the "a" that is throwing you off. What is throwing you haven't used the chain rule as Prove It said.

You have the derivative of $\sqrt{2a^3x- x^4}= \left(2a^3x- x^4\right)^{1/2}$ as $\frac{1}{2}\left(2a^3x- x^5\right)^{-1/2}$. That's wrong because, by the chain rule, you must multiply that by the derivative of $2a^x- x^4$ with respect to x.

After some trying..

$\dfrac{\dfrac{a^3-2x^3}{(2a^3x-x^4)^{1/2}}-{\dfrac{a^3}{(3a^2x)^{2/3}}}}{-\dfrac{3ax^2}{4(ax^3)^{3/4}}}$

Is this closer?