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

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Discussion Overview

The discussion revolves around evaluating the limit of the expression (cos(x)/x^2) - (sin(x)/x^3) as x approaches 0. Participants explore the use of L'Hôpital's rule and alternative approaches to handle the limit, while also addressing the conditions necessary for applying the rule.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the applicability of L'Hôpital's rule, noting that the first part of the expression (cos(x)/x^2) does not yield a 0/0 form as cos(x) approaches 1 when x approaches 0.
  • Another participant suggests rewriting the expression to achieve a 0/0 form by combining the fractions, proposing that the limit can be expressed as (x cos(x) - sin(x))/x^3.
  • A participant expresses confusion about the rearrangement process used to combine the fractions.
  • There is a brief exchange regarding the correct spelling of L'Hôpital's name, with participants correcting each other on the modern spelling.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to evaluate the limit. Some agree on the need to rewrite the expression, while others question the initial application of L'Hôpital's rule.

Contextual Notes

There is uncertainty regarding the conditions under which L'Hôpital's rule can be applied, particularly in relation to the form of the limit as x approaches 0. The discussion also reflects varying levels of familiarity with calculus techniques among participants.

Who May Find This Useful

Students seeking assistance with limit evaluation techniques, particularly those involving L'Hôpital's rule and algebraic manipulation of expressions.

nicholas1504
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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
 
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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:

[tex]\frac{\cos x}{x^2} - \frac{\sin x}{x^3}[/tex]

As:

[tex]\frac{x \cos x - \sin x}{x^3}[/tex]

Does that help?
 
i wasn't sure which forum, i should post my query in.

How did you rearrange it??
 
a/(c^2) - b/(c^3)
= ac/(c^3) - b/(c^3)
= (ac-b)/(c^3)

-- AI
 
Thank you guys, you just made my day beautiful!
 
wasn't sure which forum, i should post my query in.

Homework help.
 
Actually, the modern spelling is l'Hôpital; the 's' is swallowed into the circumflex.
 
Right !
I love the flavor of the past.
 

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