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  • <img alt="\mathrm{ \operatorname{Lim}_{x \rightarrow 0} \frac{x \cos x-\log (1+x)}{x^2} \, \, equals}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Coperatorname%7BLim%7D_%

\mathrm{ \operatorname{Lim}_{x \rightarrow 0} \frac{x \cos x-\log (1+x)}{x^2} \, \, equals}

Option: 1

1


Option: 2

-1


Option: 3

1/2


Option: 4

-1/2


Answers (1)

best_answer

\mathrm{ \operatorname{Lim}_{x \rightarrow 0} \frac{x \cos x-\log (1+x)}{x^2} }

                                                           [ use series expansion]

\mathrm{=\lim _{x \rightarrow 0} \frac{x\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}+\ldots . .\right)-\left(x-\frac{x^2}{2}+\frac{x^3}{3}+\ldots \ldots\right)}{x^2}}

\mathrm{ =\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^3}{2 !}+\frac{x^4}{4}+\ldots \ldots}{x^2} }

                                                                                    (Apply L'Hospital's Rule)

\mathrm{=\operatorname{Lim}_{x \rightarrow 0} \frac{1}{2}-\frac{x}{3}-\frac{x}{2 !}+\frac{x^2}{4}+\ldots}

                                                                                   (Apply L'Hospital's Rule)

=\frac{1}{2}

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