\(\lim\limits_{\text x \to0}\cfrac{tan\,2\text x-sin\,2\text x}{\text x^3} \)

lim(x→0) (tan 2x - sin 2x)/x³

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

  In this Case, indeterminate Form is \(\cfrac00\) 

 \(\underset{x\to0}{lim}\frac{tan\,2x-sin\,2x}{x^3}\) \(=\underset{x\to0}{lim}\frac{sin\,2x-sin\,2x\,cos\,2x}{x^3}\)  (Multiplying numerator by cos 2x because \(\underset{x\to0}{lim}\) cos 2x = 1)

\(=\underset{x\to0}{lim}\frac{sin\,2x(1-cos\,2x)}{x^3}\) \(=\underset{x\to0}{lim}\frac{2sin\,2x}{2x}\times\frac{4(1-cos\,2x)}{(2x)^2}=4\)

Therefore, \(\underset{x\to0}{lim}\frac{tan\,2x-sin\,2x}{x^3}=4\)

Other method: - \(\underset{x\to0}{lim}\) \(\frac{tan\,2x-sin\,2x}{x^3}\) \((\frac00)\)

\(=\underset{x\to0}{lim}\) \(\frac{2sec^2\,2x-2cos\,2x}{3x^2}\) \((\frac00)\)

\(=\underset{x\to0}{lim}\) \(\frac{8sec^22x\,tan\,2x+4\,sin\,2x}{6x}\)

\(=\frac83\,\underset{x\to0}{lim}\,sec^22x\,\underset{x\to0}{lim}\frac{tan\,2x}{2x}+\frac{4}{3}\,\underset{x\to0}{lim}\frac{sin\,2x}{2x}\)

\(=\frac{8}{3}+\frac{4}{3}=\frac{12}{3}=4\)