# Find the limit lim_{xrightarrow-infty}frac{x^2-4x+8}{3x^3}

Question
Limits and continuity
Find the limit
$$\lim_{x\rightarrow-\infty}\frac{x^2-4x+8}{3x^3}$$

2021-02-26
The given function is
$$\lim_{x\rightarrow-\infty}\frac{x^2-4x+8}{3x^3}$$
Solve the limit of the function
$$=\frac{1}{3}\lim_{x\rightarrow-\infty}\frac{x^2-4x+8}{x^3}$$
$$=\frac{1}{3}\lim_{x\rightarrow-\infty}(\frac{1}{x}-\frac{4}{x^2}+\frac{8}{x^3})$$
$$=\frac{1}{3}\lim_{x\rightarrow-\infty}[\lim_{x\rightarrow-\infty}(\frac{1}{x})-\lim_{x\rightarrow-\infty}(\frac{-4}{x^2})+\lim_{x\rightarrow-\infty}(\frac{8}{x^3})]$$
$$=\frac{1}{3}(0+0+0)$$
$$=0$$
Hence, the limit of the given function is 0.

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