Lim((cosx/x^2)-(sinx)/x^3)) x->0 L'hopital

[nicholas1504]

Hey Everybody.

I was just wondering about this query:

lim((cosx/x^2)-(sinx)/x^3))
x->0

My teacher is telling me to use L'hopital, but the problem is that

the first part (cosx)/(x^2) isn't a "0/0", cosx-> 1 for x->0

So what should I do, i know the right answer is -1/3, but i need to prove it.

Plz. Help Me

Thanks

 

[Zurtex]

I've never used that method in calculus before, but am I right in thinking that you need to get to a 0/0 situation?

In which it seems fairly obvious to me that you should rewrite:

\frac{\cos x}{x^2} - \frac{\sin x}{x^3}

As:

\frac{x \cos x - \sin x}{x^3}

Does that help?

 

[nicholas1504]

i wasn't sure which forum, i should post my query in.

How did you rearrange it??

 

[TenaliRaman]

a/(c^2) - b/(c^3)
= ac/(c^3) - b/(c^3)
= (ac-b)/(c^3)

-- AI

 

[nicholas1504]

Thank you guys, you just made my day beautiful!

 

[Hurkyl]

wasn't sure which forum, i should post my query in.

Homework help.

 

[humanino]

Regle du marquis Guillaume de l'Hospital

I can't help correcting french name misspells

 

[Hurkyl]

Actually, the modern spelling is l'Hôpital; the 's' is swallowed into the circumflex.

 

[humanino]

Right !
I love the flavor of the past.