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# Help me on this

2 months, 2 weeks ago

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

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 $$\displaystyle \lim_{x\to\infty} \left(e^{3x}-x\right)^{\dfrac{1}{4x}}=e^{3/4}$$ · 2 months, 2 weeks ago