The discussion revolves around evaluating limits of the form (0/0)^infinity, specifically through the lens of L'Hôpital's Rule and related techniques. Participants explore the nuances of applying L'Hôpital's Rule to a limit involving an exponential expression.
The conversation is active, with various approaches being considered. Some participants suggest that taking the logarithm of the expression may lead to a clearer path for evaluation, while others emphasize the relevance of L'Hôpital's Rule as indicated in the thread title. There is no explicit consensus on the best approach yet.
Participants are navigating the complexities of limits involving indeterminate forms and are considering the implications of different mathematical techniques. The discussion reflects a mix of familiarity with the topic and uncertainty about the most effective methods to apply.
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.