Can L'Hopital's Rule be used to solve (0/0)^infinity limits?

In summary: However, I would do it in a more simplified way by using the fact that$$\frac{e^x + e^{-x} - 2}{x^2} = \frac{(e^{x/2} - e^{-x/2})^2}{x^2}.$$In summary, L'Hôpital needs to calculate the derivatives of numerator and denominator. It may need more effort and a few Taylor expansions to do so. If it does, the final answer will be obtained by exponentiating.
  • #1
Crystal037
167
7
Homework Statement
lim x tends to 0 (((e^x+e^(-x)-2)/(x^2))^(1/x^2))
Relevant Equations
lim x tends to 0 (e^x-1)/x=1
It is of the form (0/0)^infinity. I know how to solve 1^infinity form
 
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  • #2
Is it ##\lim_{x \to 0} \left( \dfrac{e^x+e^{-x}-2}{x^2}\right)^{\frac{1}{x^2}}\,?##

For L'Hôpital you will have do calculate the derivatives of numerator and denominator.
 
  • #3
Yes but should i consider the power 1/x^2 while differentiating the numerators and denominator
 
  • #4
I would take it for both: ##\left( e^x+e^{-x} - 2\right)^{x^{-2}}## and ##\left( x^2 \right)^{x^{-2}}##. If I'm right, then the limit without the power of ##\frac{1}{x^2}## is ##1##, but with it, it isn't. But I'm not sure whether L'Hôpital will do. Maybe it needs more effort and a few Taylor expansions.
 
  • #5
fresh_42 said:
I would take it for both: ##\left( e^x+e^{-x} - 2\right)^{x^{-2}}## and ##\left( x^2 \right)^{x^{-2}}##. If I'm right, then the limit without the power of ##\frac{1}{x^2}## is ##1##, but with it, it isn't. But I'm not sure whether L'Hôpital will do. Maybe it needs more effort and a few Taylor expansions.
I wouldn't do things this way. Instead, I would let ##y = \left(\frac{ e^x+e^{-x} - 2}{x^2}\right)^{1/x^2}##, take ln of both sides, and then take the limit. The result you get, if a limit exists will be the limit of the ln of that expression, so the final answer is obtained by exponentiating.

I haven't worked the problem, but this is the usual approach when dealing with limits with some expression raised to a variable power.
 
  • #6
Yes, but L'Hôpital was given in the thread title.
 
  • #7
fresh_42 said:
Yes, but L'Hôpital was given in the thread title.
The limits after taking the log can be evaluated using the Hospital rule. According to Mathematica, the limit is equal to ##e^{1/12}##.

You can simplify the algebra by using the fact that
$$\frac{e^x + e^{-x} - 2}{x^2} = \frac{(e^{x/2} - e^{-x/2})^2}{x^2} = \left[\frac{\sinh (x/2)}{x/2}\right]^2.$$ I had to apply L'Hopital's rule several times and eventually arrived at the correct result.
 
  • #8
fresh_42 said:
Yes, but L'Hôpital was given in the thread title.
My response didn't preclude using L'Hopital's Rule in evaluating the limit.
 

FAQ: Can L'Hopital's Rule be used to solve (0/0)^infinity limits?

What are limits and how do they relate to L'hopital's rule?

Limits refer to the value that a function approaches as its input approaches a certain value. L'hopital's rule is a mathematical tool used to evaluate certain types of limits by taking the derivative of the numerator and denominator separately.

When should L'hopital's rule be used?

L'hopital's rule should be used when evaluating a limit that results in an indeterminate form, such as 0/0 or ∞/∞. It can also be used to simplify complicated limits involving trigonometric or exponential functions.

What is the process for using L'hopital's rule?

The process for using L'hopital's rule involves taking the derivative of both the numerator and denominator of the original limit, simplifying if possible, and then evaluating the new limit. This process can be repeated if necessary until the limit can be evaluated without using L'hopital's rule.

Are there any limitations to using L'hopital's rule?

Yes, there are limitations to using L'hopital's rule. It can only be used when the limit is in an indeterminate form, and it may not work for all types of functions. Additionally, it should be used with caution and other methods should be considered when evaluating limits.

Can L'hopital's rule be used for one-sided limits?

Yes, L'hopital's rule can be used for one-sided limits. The process is the same as for two-sided limits, but the derivative is only taken in the direction of the limit being evaluated. For example, if evaluating a limit from the left, only the left-hand derivative is taken.

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