Solving Limit of (x^2+sinx)/x^2: L'Hopital's Rule Not Working

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Discussion Overview

The discussion revolves around the limit of the expression (x^2 + sin(x)) / x^2 as x approaches infinity, specifically addressing why L'Hopital's Rule may not be applicable or useful in this context. Participants explore the conditions under which L'Hopital's Rule can be applied and alternative methods for evaluating the limit.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about why L'Hopital's Rule does not seem to work for the limit as x approaches infinity of (x^2 + sin(x)) / x^2.
  • One participant suggests that while L'Hopital's Rule can be applied once, it does not yield a useful result because the limit of the derivative of the numerator does not exist as x approaches infinity.
  • Another participant argues that the limit of the original expression exists and equals 1, despite the limit of the derivatives not existing.
  • Some participants propose that the expression can be simplified to 1 + (sin(x) / x^2) and suggest using the squeeze theorem to evaluate the limit.
  • There is contention regarding the applicability of L'Hopital's Rule, with some asserting that the conditions for its use are not met due to the behavior of the derivatives.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the applicability of L'Hopital's Rule in this case. There are competing views on whether the rule can be applied and whether the limit of the derivatives is relevant to the evaluation of the limit of the original expression.

Contextual Notes

Some participants note that the limit of the derivative of the numerator does not exist, which raises questions about the conditions for applying L'Hopital's Rule. There are also discussions about the validity of the limit of the original expression and the use of alternative methods like the squeeze theorem.

ookt2c
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a challenge problem in my book states why dosent lhopitals rule work on the limit as x goes to infinity of (x^2+sinx)/x^2. I am stumped
 
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ookt2c said:
a challenge problem in my book states why dosent lhopitals rule work on the limit as x goes to infinity of (x^2+sinx)/x^2. I am stumped

I don't know if it doesn't work, it just isn't useful since differentiating once does not get you any further ahead and differentiating twice is not valid because the second derivative of the denominator is not defined for any open interval containing infinity.

http://en.wikipedia.org/wiki/L'Hôpital's_rule

To correctly way to solve the limit is expand the above as follows

(x^2+sinx)/x^2=1+(sinx)/x^2

bound the value of sinx and then use the squeeze theorem to show the limit is equal to one.
 
ookt2c said:
a challenge problem in my book states why dosent lhopitals rule work on the limit as x goes to infinity of (x^2+sinx)/x^2. I am stumped

It is only valid to use L'Hopital's rule once on this function, as you have both numerator and denominator increasing without bound. After you get the expression 1 + [itex]\lim_{x\rightarrow\infty} \frac{\cos x}{2x}[/itex], the expression no longer conforms to the conditions that validate the use of L'Hopital's rule.
 
No, slider.

One of the conditions for the applicability of Hospital's rule is that the limit of the fraction of the derivatives MUST exist.

In this case, that limit does not exist, but the limit of our original expression exist NONETHELESS (equaling 1).

Read John's link carefully.
 
arildno said:
No, slider.

One of the conditions for the applicability of Hospital's rule is that the limit of the fraction of the derivatives MUST exist.

In this case, that limit does not exist, but the limit of our original expression exist NONETHELESS (equaling 1).

Read John's link carefully.

If you apply it once the limit still exists.
 
John Creighto said:
If you apply it once the limit still exists.
Which limit? The limit of the original problem certainly exists but arildno's point is that trying to apply L'Hopital's rule the derivative of the numerator is 2x+ cos(x) and that has no limit as x goes to infinity.
 
HallsofIvy said:
Which limit? The limit of the original problem certainly exists but arildno's point is that trying to apply L'Hopital's rule the derivative of the numerator is 2x+ cos(x) and that has no limit as x goes to infinity.

Yes but the denominator is x^2 so you can still apply the rule once and the limit will still exist.
 
?? Now I am starting to wonder if you know what "L'Hopital's" rule is? As I said before, the limit of the derivative of the numerator does not exist. It doesn't matter what the denominator is!
 
arildno said:
No, slider.

One of the conditions for the applicability of Hospital's rule is that the limit of the fraction of the derivatives MUST exist.

In this case, that limit does not exist, but the limit of our original expression exist NONETHELESS (equaling 1).

Read John's link carefully.

Ah yes. I was asleep when I wrote that nonsense.
 

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