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

• ookt2c
In summary: Sorry about that.In summary, L'Hopital's rule does not work for the limit as x goes to infinity of (x^2+sinx)/x^2 because the limit of the derivative of the numerator does not exist, one of the conditions for the rule to be applicable. The correct way to solve this limit is to expand the expression and use the squeeze theorem.
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

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

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).

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).

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).

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

What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical technique used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the ratio of two functions is indeterminate, then the limit of the ratio of their derivatives will be the same.

Why is L'Hopital's Rule not working for this limit?

There are a few possible reasons why L'Hopital's Rule may not be working for a specific limit. One common reason is that the limit may not be in an indeterminate form, and therefore L'Hopital's Rule does not apply. Another reason could be that the derivatives of the functions involved do not exist at the point where the limit is being evaluated.

Are there any other methods for solving this limit?

Yes, there are other methods for solving limits besides L'Hopital's Rule. Some common techniques include substitution, factoring, and using trigonometric identities. It may also be helpful to graph the functions involved to gain a better understanding of the limit.

Can you provide an example of when L'Hopital's Rule would not work for a limit?

One example is the limit as x approaches 0 of (x^2+sinx)/x^2. In this case, applying L'Hopital's Rule would result in a limit of 1, but the actual limit is 2. This is because the limit is not in an indeterminate form, and the derivatives of the numerator and denominator do not exist at x=0.

Is L'Hopital's Rule always a reliable method for solving limits?

No, L'Hopital's Rule is not always a reliable method for solving limits. It should only be used when the limit is in an indeterminate form and the derivatives of the functions involved exist at the point where the limit is being evaluated. It is always important to double check the result obtained using L'Hopital's Rule to ensure its accuracy.

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