Question 37

Suppose

,$$x \epsilon \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ .Then $$\lim_{x \rightarrow 0} \frac{f(x)}{x^2}$$

Solution

f(x) = $$cosx (x^2*2-x*2x) - x(2sinx *2-2x*tanx) + 1(2sinx *x-tanx*x^2)$$

= $$x^2\tan x-2x\sin x$$

$$\lim_{x \rightarrow 0} \frac{f(x)}{x^2}$$

$$=\lim_{x\rightarrow0}\frac{x^2\tan x}{x^2}-\frac{\left(2x\sin x\right)}{x^2}$$

=$$\lim_{x\rightarrow0}\tan x-\frac{2\sin x}{x}$$

= 0-2=-2

B is the correct answer.

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