Question

Lim x-0 (sin2xcsc3xsec2x)/x²cot²4x

Answer 1

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

`(\sin2x\csc3x\sec2x)/(x^2\cot^24x)=(\sin2x\sin^24x)/(x^2\sin3x\cos2x\cos^24x)`

Then we rewrite the fraction as

`(\sin2x)/(2x)\left((\sin4x)/(4x)\right)^2(3x)/(\sin3x)(32)/(3\cos2x\cos^24x)`

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

`\displaystyle\lim_(x\to0)(\sin ax)/(ax)=\lim_(x\to0)(ax)/(\sin ax)=1`

for
`a\\eq0`. So when we take the limit, we have

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

`=1\cdot1^2\cdot1\cdot\frac{32}3=\frac{32}3`