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

[ookt2c]

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

 

[John Creighto]

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.

 

[slider142]

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 + $\lim_{x\rightarrow\infty} \frac{\cos x}{2x}$, the expression no longer conforms to the conditions that validate the use of L'Hopital's rule.

 

[arildno]

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.

 

[John Creighto]

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.

 

[HallsofIvy]

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.

 

[John Creighto]

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.

 

[HallsofIvy]

?? 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!

 

[slider142]

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.