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- 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
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.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.
The limits after taking the log can be evaluated using the Hospital rule. According to Mathematica, the limit is equal to ##e^{1/12}##.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.fresh_42 said:Yes, but L'Hôpital was given in the thread title.