Question

Evaluate : limx→ tan2-sin2x x3

Answer 1

Given:

`\lim _(x\to0)(\tan 2x-\sin 2x)/(x^3)`

Solve:

`\lim _(x\to0)(\tan 2x-\sin 2x)/(x^3)`

Use l'hopital's rule:

`\begin{gathered} =\lim _(x\to0)((d)/(dx)(-\sin 2x+\tan 2x))/((d)/(dx)(x^3)) \\ =\lim _(x\to0)(-2\cos (2x)+2\tan ^2(2x)+2)/(3x^2) \end{gathered}`

Simplify:

`\begin{gathered} =\lim _(x\to0)(-2\cos (2x)+2\tan ^2(2x)+2)/(3x^2) \\ =\lim _(x\to0)(2(-\cos (2x)+\tan ^2(2x)+1))/(3x^2) \end{gathered}`

Apply the constant multiple rule:

`\begin{gathered} \lim _(x\to0)cf(x)=c\lim _(x\to0)f(x) \\ \text{With c=}(2)/(3) \\ f(x)=(-\cos (2x)+\tan ^2(2x)+1)/(x^2) \end{gathered}`
`\begin{gathered} =(2\lim _(x\to0)(-\cos (2x)+\tan ^2(2x)+1)/(x^2))/(3) \\ =(2\lim _(x\rightarrow0)((4\tan ^2(2x)+4)\tan (2x)+2\sin (2x))/(2x))/(3) \end{gathered}`

Similary :

`\begin{gathered} =(2\lim _(x\to0)(2\cos (2x)+12\tan ^4(2x)+16\tan ^2(2x)+4))/(3) \\ =(2(6))/(3) \\ =4 \end{gathered}`