$$\lim_{x\rightarrow 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$ I attempted to expand $\sin x$ and am unsure how to proceed. $$1-\frac{x}{3!}\left[x^2-\frac{x^4}{5\times 4} \cdots \right]$$
Can anyone please help me with this Problem?
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2Does this answer your question? Find a limit in an efficent way - found using an Approach0 search. There's also other duplicates, e.g., Evaluate $\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$. – John Omielan Sep 27 '23 at 01:07
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Note that all $4$ answers of the proposed duplicate, and all but one answer among the $5$ of the other linked question, use techniques other than L'Hôpital's rule (based on your implied request using the "limits-without-lhopital" tag). Also, both questions only deal with $x\to 0^{+}$, but you can use that result to get what the limit for $x\to 0^{-}$ would be. Last, but not least, welcome to Math SE. – John Omielan Sep 27 '23 at 01:16
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1If you're interested in more examples like this, see the 2nd half of this 28 December 2007 sci.math post. In case anyone is wondering why the OP's limit isn't one of the 24 numbered limit examples I gave, that's because the OP's limit was one of 3 or 4 such limits (exponential indeterminate forms) that were carefully worked out before I listed those additional 24 limits for students to practice on, this being in the handout of mine (written back in the late 1990s) that I copied those 24 examples from. – Dave L. Renfro Sep 27 '23 at 08:40
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Write it in e to the power ln format . Then u get 1/x(Ln(sin/x)) then put limit for sin/x. I think u got the idea
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This needs work to be an actual answer. It’s fine (although hard to decipher) as a comment. – Ted Shifrin Sep 27 '23 at 02:55
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Let $f(x) = \lim_{x\to0} (\frac{\sin(x)}{x})^{1/x}$.
A nice property of logarithms is that they can bring an exponent within a function in front. Because of this, we have $$\ln(f(x)) = \frac{\ln(\lim_{x\to0}\frac{\sin(x)}{x})}{x}$$
We know from the squeeze theorem that $$\lim_{x\to0}\frac{\sin(x)}{x} = 1$$ Thus, we have $$\ln(f(x)) = \frac{\ln(1)}{x}$$
$\ln(1) = 0$, so we have $$\ln(f(x)) = \frac{0}{x} = 0$$
Following this, we know that $$\ln(f(x)) = 0$$ so we take both sides to the power of e: $$e^{\ln(f(x))} = e^0$$ $$f(x) = 1$$
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2The formula for $f(x)$ does not make sense, as the right hand side does not depend on $x.$ If you consider the limit of $\left ({\sin x\over x}\right )^{1/x^2}$ your method will give a false result. – Ryszard Szwarc Sep 27 '23 at 14:46