- #1

Crystal037

- 154

- 5

- 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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- Thread starter Crystal037
- Start date

- #1

Crystal037

- 154

- 5

- 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

- #2

- 17,242

- 17,235

For L'Hôpital you will have do calculate the derivatives of numerator and denominator.

- #3

Crystal037

- 154

- 5

Yes but should i consider the power 1/x^2 while differentiating the numerators and denominator

- #4

- 17,242

- 17,235

- #5

Mark44

Mentor

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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

- 17,242

- 17,235

Yes, but L'Hôpital was given in the thread title.

- #7

vela

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

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The limits after taking the log can be evaluated using the Hospital rule. According to Mathematica, the limit is equal to ##e^{1/12}##.Yes, but L'Hôpital was given in the thread title.

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

Mark44

Mentor

- 36,330

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My response didn't preclude using L'Hopital's Rule in evaluating the limit.Yes, but L'Hôpital was given in the thread title.

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