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  • <img alt="\mathrm{\text { calculate } \lim _{x \rightarrow 0} \frac{[\sin x-x][\cos (3 x)-1]}{x\left[e^x-1\right]^4}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext%20%7B

\mathrm{\text { calculate } \lim _{x \rightarrow 0} \frac{[\sin x-x][\cos (3 x)-1]}{x\left[e^x-1\right]^4}}

Option: 1

2 / 3


Option: 2

3 / 4


Option: 3

3


Option: 4

1


Answers (1)

best_answer

Very rough analysis with Taylor series: \mathrm{\sin x-x } vanishes to third order and \mathrm{\cos (3 x)-1} vanishes to second order. The denominator vanishes to fifth order, so the limit should be a nonzero real number.
To be much more precise,

\mathrm{\begin{aligned} \sin x-x & =-\frac{1}{6} x^3+O\left(x^5\right) \\ \cos (3 x)-1 & =-\frac{(3 x)^2}{2}+O\left(x^4\right) \\ e^x-1 & =x+O\left(x^2\right) \end{aligned}}

Convince yourself that all these high order terms can indeed be neglected, and then compute 

\mathrm{ (-1 / 6)\left(-(3 / 2)^2\right)=3 / 4}

Posted by

Irshad Anwar

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