Question

Lim x-0 (sinxcos2xcosx)/sec4xtan3x

Answer 1

Same general strategy as with your other question: rewrite the fraction to include factors of
`(\sin ax)/(ax)` or
`(ax)/(\sin ax)`.

`(\sin x\cos2x\cos x)/(\sec4x\tan3x)=(\sin x\cos x\cos2x\cos3x\cos4x)/(\sin3x)`

`\implies\displaystyle\lim_(x\to0)\cdots=\frac13\lim_(x\to0)\frac{\sin x}x\lim_(x\to0)(3x)/(\sin3x)\lim_(x\to0)\cos x\cos2x\cos3x\cos4x`

All limits approach 1, so the ultimate limit is
`\frac13`.