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I am to find the limit of $$\lim_{x \to \infty} \left(1+\frac{x}{5x^3+x^2+8}\right)^ {\dfrac{x^3+8}{x}}$$

I could not find the proper substitution here. I would be happy if someone could shed some light. Thanks.

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It looks like $$\lim_{x\to\infty}(1+\frac 1{5x^2})^{x^2}$$ for large x, if that technique is correct... –  abiessu Nov 25 '13 at 22:38

2 Answers 2

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$$ \begin{align} \lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^{\frac{x^3+8}{x}} &=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{x^2+8/x}\\ &=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2+x+8/x)\frac{x^2+8/x}{5x^2+x+8/x}}\\ &=\left(\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{(5x^2+x+8/x)}\right)^{\lim\limits_{x\to\infty}\frac{x^2+8/x}{5x^2+x+8/x}}\\[6pt] &=e^{1/5} \end{align} $$

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$$\lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^ {x^2+8/x}=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^ {x^2+8/x}=$$ $$=\lim_{x\to\infty}\left(\left(1+\frac{1}{5x^2+x+8/x}\right)^{5x^2+x+8/x} \right)^{\frac{x^2+8/x}{5x^2+x+8/x}}=e^{1/5}$$ because $$\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{5x^2+x+8/x}=e$$ and $$\lim_{x\to\infty}{\frac{x^2+8/x}{5x^2+x+8/x}}=\frac{1}{5}$$

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