Respuesta :
For starters, we can write
[tex]\cos^3x-\sin^3x=(\cos x-\sin x)(\cos^2x+\cos x\sin x+\sin^2x)=(\cos x-\sin x)(1+\cos x\sin x)[/tex]
Then we can absorb the factor of [tex]\cos x[/tex] from the denominator into the numerator:
[tex]\dfrac{x(\cos^3x-\sin^3x)}{\tan4x\cos x}=\dfrac{x(1-\tan x)(1+\cos x\sin x)}{\tan 4x}[/tex]
Then provided that
[tex]\displaystyle\lim_{x\to\pi/4}\frac{1-\tan x}{\tan4x}[/tex]
exists, we can split the limit of the product into the product of limits:
[tex]\displaystyle\lim_{x\to\pi/4}\frac{x(\cos^3x-\sin^3x)}{\tan4x\cos x}=\left(\lim_{x\to\pi4}x(1+\cos x\sin x)\right)\left(\lim_{x\to\pi4}\dfrac{1-\tan x}{\tan 4x}\right)[/tex]
[tex]x(1+\cos x\sin x)[/tex] is continuous at [tex]x=\dfrac\pi4[/tex], so that limit reduces to [tex]\dfrac\pi4\left(1+\cos\dfrac\pi4\sin\dfrac\pi4\right)=\dfrac{3\pi}8[/tex].
The remaining limit is of indeterminate form 0/0; one application of L'Hopital's rule shows that the limit would be
[tex]\displaystyle\lim_{x\to\pi/4}\frac{1-\tan x}{\tan4x}=\lim_{x\to\pi/4}\frac{-\sec^2x}{4\sec^24x}=-\dfrac12[/tex]
so the final limit would be [tex]-\dfrac{3\pi}{16}[/tex].
But if you haven't learned about L'Hopital's rule (or if you're like me and prefer doing more work for some reason), we have to try something else: more trig identities!
Euler's formula and DeMoivre's theorem are very useful here:
[tex]e^{ix}=\cos x+i\sin x[/tex] (Euler)
[tex](\cos x+i\sin x)^n=\left(e^{ix}\right)^n=e^{inx}[/tex] (DeMoivre)
Take [tex]n=4[/tex]; then
[tex]e^{4ix}=\cos4x+i\sin4x=(\cos^4x-6\cos^2x\sin^2x+\sin^4x)+i(4\cos^3x\sin x-4\cos x\sin^3x)[/tex]
[tex]\implies\begin{cases}\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x\\\sin4x=4\cos^3x\sin x-4\cos x\sin^3x\end{cases}[/tex]
So we have
[tex]\tan 4x=\dfrac{4\cos^3x\sin x-4\cos x\sin^3x}{\cos^4x-6\cos^2x\sin^2x+\sin^4x}[/tex]
On the right side, divide through the numerator and denominator by [tex]\cos^4x[/tex]; doing so yields
[tex]\tan 4x=\dfrac{4\frac{\sin x}{\cos x}-4\frac{\sin^3x}{\cos^3x}}{1-6\frac{\sin^2x}{\cos^2x}+\frac{\sin^4x}{\cos^4x}}[/tex]
[tex]\implies\tan4x=\dfrac{4\tan x-4\tan^3x}{1-6\tan^2x+\tan^4x}=\dfrac{4\tan x(1-\tan x)(1+\tan x)}{1-6\tan^2x+\tan^4x}[/tex]
So in the limit, we can simplify the rational expression even further:
[tex]\displaystyle\lim_{x\to\pi/4}\frac{1-\tan x}{\tan4x}=\lim_{x\to\pi/4}\frac{1-\tan x}{\frac{4\tan x(1-\tan x)(1+\tan x)}{1-6\tan^2x+\tan^4x}}=\lim_{x\to\pi/4}\frac{1-6\tan^2x+\tan^4x}{4\tan x(1+\tan x)}[/tex]
and this function happens to be continuous at [tex]x=\dfrac\pi4[/tex], so we can evaluate directly:
[tex]\displaystyle\lim_{x\to\pi/4}\frac{1-\tan x}{\tan4x}=\frac{1-6\tan^2\frac\pi4+\tan^4\frac\pi4}{4\tan\frac\pi4\left(1+\tan\frac\pi4\right)}=-\frac48=-\frac12[/tex]
We end up with the same limit we found earlier, [tex]-\dfrac{3\pi}{16}[/tex].
