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Question Number 194139 by cortano12 last updated on 28/Jun/23

Commented by MM42 last updated on 28/Jun/23

$$for$$$${lim}_{x\to 0}\frac{x^2+2cosx-2}{x^4}\to hop$$$${lim}_{x\to 0}\frac{2(x-sinx)}{4x^3}=\frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$$$$\Rightarrow$$

Answered by MM42 last updated on 28/Jun/23

$$hop\to {lim}_{x\to 0}\frac{2x+2sinx}{4x^3}=+\mathrm{\infty }$$$$$$

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