Question

Lim x-0 (sin2xcsc3xsec2x)/x²cot²4x

Answer 1

By the definitions of cosecant, secant, and cotangent, we have

$\dfrac{\sin2x\csc3x\sec2x}{x^2\cot^24x}=\dfrac{\sin2x\sin^24x}{x^2\sin3x\cos2x\cos^24x}$

Then we rewrite the fraction as

$\dfrac{\sin2x}{2x}\left(\dfrac{\sin4x}{4x}\right)^2\dfrac{3x}{\sin3x}\dfrac{32}{3\cos2x\cos^24x}$

The reason for this is that we want to apply the well-known limit,

$\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1$

for $a\neq0$. So when we take the limit, we have

$\displaystyle\lim_{x\to0}\cdots=\lim_{x\to0}\frac{\sin2x}{2x}\left(\lim_{x\to0}\frac{\sin4x}{4x}\right)^2\lim_{x\to0}\frac{3x}{\sin3x}\lim_{x\to0}\frac{32}3\cos2x\cos^24x}$

$=1\cdot1^2\cdot1\cdot\dfrac{32}3=\dfrac{32}3$