Question

Evaluate:
$\rm \displaystyle \lim_{x\to \pi/2}\ \frac{3\cos x+\cos 3x}{(2x-\pi)^{3}}$

Answer 1

Hope it helppsss uhhhg

Answer 2

we have to evaluate, $\rm \displaystyle \lim_{x\to \pi/2}\ \frac{3\cos x+\cos 3x}{(2x-\pi)^{3}}$

substitute , x = h + π/2

so, $\rm \displaystyle \lim_{h+\frac{\pi}{2}\to \pi/2}\ \frac{3\cos (h+\pi/2)+\cos 3(h+\pi/2)}{\{2(h+\pi/2)-\pi\}^{3}}$

= $\displaystyle\lim_{h\to 0}\frac{-3sin(h)+sin(3h)}{(2h)^3}$

we know, sin3Φ = 3sinΦ - 4sin³Φ

so, sin(3h) = 3sin(h) - 4sin³(h)

so, $\displaystyle\lim_{x\to 0}\frac{-3sin(h)+3sin(h)-4sin^3(h)}{8h^3}$

= $\frac{-1}{2}\displaystyle\lim_{x\to 0}\left(\frac{sin(h)}{h}\right)^3$

we know, $\\displaystyle\lim_{h\to 0}\frac{sinh}{h}=1$

= -1/2 × 1 = -1/2 [ Ans]