Question

HELP PLZZ!!
lim approaches x=> 0
sec^2 2x -1+sin^2 4x/ 3xsinx

Answer 1

\[\frac{ \sec ^{2} 2x -1+ \sin ^{2} 2x }{ 3xsinx }\]

Answer 2

divide top and bottom by (2x)^2

Answer 3

\[\lim_{x \rightarrow 0}\frac{\frac{\sec^2(2x)-1}{(2x)^2}+(\frac{\sin(2x)}{2x})^2}{\frac{3}{4}\frac{x}{x} \frac{\sin(x)}{x}}\]

Answer 4

A slightly different manipulation:\[\begin{align*}\frac{\sec^22x-1+\sin^22x}{3x\sin x}&=
\frac{\tan^22x+\sin^22x}{3x\sin x}\\\\
&=
\frac{\dfrac{\sin^22x}{\cos^22x}+\sin^22x}{3x\sin x}\\\\
&=
\frac{\sin^22x\left(\sec^22x+1\right)}{3x\sin x}\\\\
&=
\frac{4\sin^2x\cos^2x\left(\sec^22x+1\right)}{3x\sin x}\\\\
&=
\frac{4\sin x\cos^2x\left(\sec^22x+1\right)}{3x}\\\\
&=
\frac{\sin x}{x}\times\frac{4\cos^2x\left(\sec^22x+1\right)}{3}\end{align*}\]