1. ## L'Hopital's Rule help please

I need help with this question. I think I'm supposed to use L'Hopital's Rule, but I'm not sure if I'm doing it correctly.

lim as x -->0 (2x^6+6x^3)/(4x^5+3x^3)

I got 0 at first, but then I got 2. I'm just so unsure! Any help at all is greatly appreciated!

2. If you're going to use L'Hospital's Rule

$\displaystyle \displaystyle \lim_{x \to 0}\frac{2x^6 + 6x^3}{4x^5 + 3x^3} = \lim_{x \to 0}\frac{2x^3 + 6}{4x^2 + 3}$

$\displaystyle \displaystyle = \lim_{x \to 0}\frac{6x^2}{8x}$ by L'Hospital's Rule

$\displaystyle \displaystyle = \lim_{x \to 0}\frac{12x}{8}$ by L'Hospital's Rule

$\displaystyle \displaystyle = 0$.

3. How did you get 2 ??

4. I applied L'Hopital's rule like 4 times then stopped. I knew that would be wrong :/

But how did you figure out this part?

lim as x-->0= 2x^3+ 6/ 4x^2+3

I thought, you just went straight to finding the derivative after the original was found to be indeterminate? Please tell me how that was found.

5. simplify first =)

6. I think the idea is

$\displaystyle \displaystyle\frac{2x^6+6x^3}{4x^5+3x^3}=\left[\frac{x^3}{x^3}\right]\frac{2x^3+6}{4x^2+3}$

and the limit as x approaches zero is 2.

7. Ohhh! Okay! But then why do we stop simplifying at 2x^3/ 4x^2 ? Can't we keep going? Like get rid of all the x's? Please forgive me if this comes off as stupid, I just wanna get it down once and for all lol :/

8. or,
$\displaystyle \frac{2x^6+6x^3}{4x^5+3x^3}=\frac{x}{2}-\frac{3x-12}{8x^2+6}$
which has 2 as limit when x approaches 0.

9. Okay Raoh, thanks for that but I don't follow :/ I'm not good at manipulating math at all!

I see you and Archie Meade agree that the answer should be 2? Is 0 out of the question then? lol This seems to be a controversial question :P

10. I stupidly applied L'Hospital's Rule when it wasn't $\displaystyle \displaystyle \frac{0}{0}$ ><

Serves me right for doing maths at 6am without any sleep :P

11. Wow thanks guys!Good thing that's cleared up! It would've took me a while to figure this out -___- Hopefully one day I can do math like you guys lol

Thanks again!

12. Originally Posted by kikiya
Ohhh! Okay! But then why do we stop simplifying at 2x^3/ 4x^2 ? Can't we keep going? Like get rid of all the x's? Please forgive me if this comes off as stupid, I just wanna get it down once and for all lol :/
At

$\displaystyle \displaystyle\frac{2x^3+6}{4x^2+3}$

We do get rid of the x's !!
Now it's safe to set the remaining ones all to zero as we are beyond getting 0/0
which was the indeterminate form.

As you were asked to apply L'Hopital's Rule...
Initially, we have 0/0, so...

Differentiate numerator and denominator

$\displaystyle \displaystyle\frac{12x^5+18x^2}{20x^4+9x^2}$

Still 0/0, so differentiate again...

$\displaystyle \displaystyle\frac{60x^4+36x}{80x^3+18x}$

Still 0/0, so differentiate again..

$\displaystyle \displaystyle\frac{240x^3+36}{240x^2+18}$

It's no longer indeterminate, especially there is no potential for zero in the denominator,
so now it's safe to set the remaining x's to zero to evaluate the limit.