Limit of (sinx/x)^1/x^2 using L'Hospital's Rule?

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SUMMARY

The limit of \((\frac{\sin x}{x})^{\frac{1}{x^{2}}}\) as \(x\) approaches 0 is evaluated using L'Hospital's Rule. As \(x\) approaches 0, \(\frac{\sin x}{x}\) approaches 1, leading to the expression simplifying to \(1^{\infty}\). Applying L'Hospital's Rule correctly reveals that the limit converges to \(e^{-1/2}\). This conclusion is definitive and relies on the application of calculus principles.

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Homework Statement



Calculate the limit as x tends to 0:

(\frac{sinx}{x})^{\frac{1}{x^{2}}}

Homework Equations





The Attempt at a Solution

 
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when x ->0

sinx/x ->1
 
sorry, please see my other post. this one had a misprint so i posted it again.
 

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