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?


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

The above is based on

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.