A limit as x approaches a.

**qleeq** · February 10th 2011, 11:09 PM

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

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!

**Prove It** · February 10th 2011, 11:14 PM

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

**qleeq** · February 10th 2011, 11:18 PM

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?

**Chris L T521** · February 10th 2011, 11:18 PM

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.

**Prove It** · February 10th 2011, 11:31 PM

They are not constant functions, they are coefficients of the functions of $\displaystyle x$ , which is fine...

**qleeq** · February 10th 2011, 11:37 PM

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.

**Prove It** · February 10th 2011, 11:53 PM

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.

**qleeq** · February 11th 2011, 12:09 AM

I have a feeling I'm over-complicating this..

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

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

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

**Prove It** · February 11th 2011, 12:11 AM

You haven't done the differentiating properly... You need to use the Chain Rule for each term...

**qleeq** · February 11th 2011, 12:14 AM

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?

**Prove It** · February 11th 2011, 12:16 AM

Each term you've differentiated is incorrect...

**qleeq** · February 11th 2011, 01:14 AM

Figured just as much. I think the thing that's throwing me off is the 'a'.. how do I deal with it?

**Prove It** · February 11th 2011, 01:55 AM

$\displaystyle a$ gets treated as a constant.

**HallsofIvy** · February 11th 2011, 03:48 AM

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.

**qleeq** · February 11th 2011, 04:03 AM

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?

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