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  • What is <img alt="\mathrm{\lim _{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)?}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Cr

What is \mathrm{\lim _{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)?}

Option: 1

0


Option: 2

-1


Option: 3

1


Option: 4

3


Answers (1)

best_answer

It is because when you expand \mathrm{\cos (x)} and \mathrm{\ln (\cos (x))}, you need to consider more fourth-order term. Specifically,

\mathrm{cos (x)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}+o\left(x^{5}\right) }

Then
\mathrm{\ln (\cos (x))=-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{\left(-\frac{x^{2}}{2}+\frac{x^{4}}{24}\right)^{2}}{2}+o\left(x^{5}\right)=-\frac{x^{2}}{2}-\frac{x^{4}}{12}+o\left(x^{5}\right) }.

With the same expression for \mathrm{\sin (x) } as you have written, we obtain
\mathrm{\frac{1}{\ln (\cos (x))}+\frac{2}{\sin ^{2}(x)} =\frac{1}{-\frac{x^{2}}{2}-\frac{x^{4}}{12}+o\left(x^{5}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)} }
                                             \mathrm{=\frac{x^{2}-\frac{x^{4}}{3}-x^{2}-\frac{x^{4}}{6}}{-\frac{x^{4}}{2}+o\left(x^{4}\right)} }
                                              \mathrm{ =\frac{x^{4}}{x^{4}+o\left(x^{4}\right)} }. 
                                              \mathrm{ =1 }

Posted by

avinash.dongre

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