Evaluate lim(x→0)(tan2x-sin2x)/x^3

Question

Evaluate: \(\lim\limits_{x \to 0}\frac{tan2x-sin2x}{x^3}\)

Answer 1

\(\lim\limits_{x \to 0} \frac{tan2x-sin2x}{x^3}\)

= \(\lim\limits_{x \to 0} \frac{\frac{sin2x}{cos2x}-sin2x}{x^3}\)

= \(\lim\limits_{x \to 0} \frac{tan2x-sin2x}{x^3}\)

= \(\lim\limits_{x \to 0} \frac{sin2x-sin2x\,cos2x}{x^3}\)

= \(\lim\limits_{x \to 0} \frac{sin\,2x(1-cos\,2x)}{x^3\,cos\,2x}\)

= \(\lim\limits_{x \to 0} \frac{tan\,2x}{x}.\lim\limits_{x \to 0}\frac{2sin^2x}{x^2}\)

= \(\lim\limits_{x \to 0} \frac{tan\,2x}{x}.2\lim\limits_{x \to 0}(\frac{sinx}{x})^2\)

= 2∙1 × 2∙(1)²

^{= 4}