Question

solve lim (tan2x-sin2x)/x to power3

Answer 1

I have done it in a copy.... Hope it will help you.......And sorry for bad handwriting!!

Answer 2

Answer:

The value of $\lim_{n \to 0}\frac{ (tan2x-sin2x)}{x^3}$ is 4

Step-by-step explanation:

Given$\lim_{n \to 0}\frac{ (tan2x-sin2x)}{x^3}$

we have to solve by applying limit

$\lim_{n \to 0}\frac{ (tan2x-sin2x)}{x^3}$=$\lim_{n \to 0}\frac{ (\frac{sin2x}{cos2x}-sin2x)}{x^3}$

                          = $\lim_{n \to 0}\frac{ (sin2x-sin2x(cos2x))}{x^3(cos2x)}$

                         = $\lim_{n \to 0}\frac{sin2x(1-cos2x)}{x^3(cos2x)}$

                        = $\lim_{n \to 0}\frac{2sinx.cosx.2sin^{2}x}{x^3(cos2x)}$

                        = $4( \lim_{n \to 0} \frac{sin^{3}x}{x^3}) (\lim_{n \to 0} \frac{cosx}{cos2x})$

                        =  $4( \lim_{n \to 0} (\frac{sin^x}{x}))^3 (\lim_{n \to 0} \frac{cosx}{cos2x})$

                       = $4(1)^3(1)$

Hence, the value of $\lim_{n \to 0}\frac{ (tan2x-sin2x)}{x^3}$ is 4