Solution: Whenever you have something raised to the power something and it looks pretty awkward, the first thing to do is to take logarithms. If you denote the limit by \(A\) then we have,

The last two lines use that fact that \(\displaystyle \lim_{x\to\infty}e^{-3x}=0\) & \(\displaystyle \lim_{x\to\infty}\dfrac{x}{e^{3x}}=\lim_{x\to\infty}\dfrac{3}{e^{3x}}=0\) by L-Hopital's rule

The answer is hence \(\displaystyle \lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}=e^{3/4}\)

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TopNewest\(\displaystyle \lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}\) :

Solution:Whenever you have something raised to the power something and it looks pretty awkward, the first thing to do is to take logarithms. If you denote the limit by \(A\) then we have,\(\begin{align} \log A &= \log\left(\lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}\right) \\ &= \lim_{x\to\infty} \log\left(e^{3x}-x\right)^{\dfrac{1}{4x}} \\ &= \lim_{x\to\infty}\dfrac{1}{4x}\log(e^{3x}-x) \\ &= \lim_{x\to\infty} \dfrac{3(e^{3x}-1)}{4(e^{3x}-x)} \quad [\text{Using L-Hopital's Rule}] \\ &= \lim_{x\to\infty} \dfrac{3(1-e^{-3x})}{4(1-xe^{-3x})} \\ &= \dfrac{3}{4}\dfrac{1-\lim_{x\to\infty}e^{-3x}}{1-\lim_{x\to\infty}xe^{-3x}} \\ &=\dfrac{3}{4}\end{align}\)

The last two lines use that fact that \(\displaystyle \lim_{x\to\infty}e^{-3x}=0\) & \(\displaystyle \lim_{x\to\infty}\dfrac{x}{e^{3x}}=\lim_{x\to\infty}\dfrac{3}{e^{3x}}=0\) by L-Hopital's rule

The answer is hence \(\displaystyle \lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}=e^{3/4}\)

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