A calculus problem by Aakhyat Singh

use expansion as tanx =x+{x^3}/3+{x^5}(2/15)+..... . and sinx=x-x^3/3!+x^5/5!+... note here in question we have *2x instead of x * so replace it to get ans. it will be 4

$\lim_{x\rightarrow0}\dfrac{\tan2x-\sin2x}{x^3}$

Both numerator and denominator go to zero, so L'Hospital's rule may be applied - three times, actually.

$\lim_{x\rightarrow0}\dfrac{\tan2x-\sin2x}{x^3}=\lim_{x\rightarrow0}\dfrac{2\sec^22x-2\cos2x}{3x^2}=\lim_{x\rightarrow0}\dfrac{8\sec^22x\tan2x+4\sin2x}{6x}=$ $\lim_{x\rightarrow0}\dfrac{32\sec2x\tan^22x+16\sec^42x+8\cos2x}{6}=\dfrac{0+16+8}6=\boxed4$