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lutik1710 [3]
1 year ago
9

Evaluate : limx→ tan2-sin2x x3

Mathematics
1 answer:
GrogVix [38]1 year ago
8 0

Given:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Solve:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Use l'hopital's rule:

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

Simplify:

\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{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=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}

Similary :

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

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