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\mathrm{\text { Evaluate } \lim _{x \rightarrow 0} \frac{1-(\cos 2 x)^3(\cos 5 x)^5(\cos 7 x)^7(\sec 4 x)^9(\sec 6 x)^{11}}{x^2}}

Option: 1

-30


Option: 2

10


Option: 3

0


Option: 4

20


Answers (1)

best_answer

Hint: Try a Taylor series expansion around 0 , and notice you can throw away many terms.

                       \mathrm{ \begin{aligned} & \lim _{x \rightarrow 0} \frac{1-(\cos 2 x)^3(\cos 5 x)^5(\cos 7 x)^7(\sec 4 x)^9(\sec 6 x)^{11}}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{1-\left(1-2 x^2\right)^3\left(1-\frac{25}{2} x^2\right)^5\left(1-\frac{49}{2} x^2\right)^7\left(1+8 x^2\right)^9\left(1+18 x^2\right)^{11}}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{1-\left(1-6 x^2\right)\left(1-\frac{125}{2} x^2\right)\left(1-\frac{343}{2} x^2\right)\left(1+72 x^2\right)\left(1+198 x^2\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{1-\left(1+x^2\left(-6-\frac{125}{2}-\frac{343}{2}+72+198\right)\right)}{x^2} \\ & =-\left(-6-\frac{125}{2}-\frac{343}{2}+72+198\right) \\ & =-30 \end{a}

I used the facts that \mathrm{\sec x=1+\frac{1}{2} x^2+\mathcal{O}\left(x^4\right), \cos x=1-\frac{1}{2} x^2+\mathcal{O}\left(x^4\right) \, and \, (1+t)^n=1+t n+\mathcal{O}\left(t^2\right).}

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Divya Prakash Singh

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