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I have to find the limit of $(3^x-x)^{1/(4x)}$ as $x\to+\infty$ without using de l'Hôpital's method or Taylor series. I've tried to use some notable limits as $(1+1/t)^t$ or other but the problem is the fact that $x$ goes to infinity, then I tried to use substitution but I can't manage it. Please can somebody help me in a clear way?

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$$f(x)=\left(3^x-x\right)^{1/4x}=\sqrt[4]3\left(1-\frac x{3^x}\right)^{1/4x}\xrightarrow[x\to\infty]{}\sqrt[4]3$$

The above is based on

$$\begin{cases}\lim_{x\to x_0}f(x)=1\\\text{and}\\\lim_{x\to x_0}g(x)=0\end{cases}\;\implies\;\lim_{x\to x_0}f(x)^{g(x)}=1$$

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