Question

limit X tends to zero sin x into 1 minus Cos 2 X upon x cube​

Answer 1

Question :-

$\to \: \tt \: \lim_{x \to 0} \frac{ \sin x(1 - \cos2x)}{ {x}^{3} }$

Answer :-

$\to \boxed{ \tt \: \lim_{x \to 0} \frac{ \sin x(1 - \cos2x)}{ {x}^{3} } = 2} \:$

Used formula :-

$\boxed{\star} \tt \: \: \: 1 - \cos2x = 2 { \sin}^{2} x \\ \\ \boxed{ \star} \: \tt \: \lim \: \frac{ \sin h}{h} = 1$

Solution :-

We have ,

$\to \: \tt \: \lim_{x \to 0} \frac{ \sin x(1 - \cos2x)}{ {x}^{3} } \\ \: \\ \because \tt 2 { \sin}^{2} x = 1 - \cos 2x \\ \\ \to \: \: \tt \: \lim_{x \to 0} \frac{ \sin x(2 { \sin}^{2} x)}{ {x}^{3} } \\ \\ \to \: \tt \: \lim_{x \to 0} \frac{ 2 { \sin}^{3}x }{ {x}^{3} } \\ \: \\ \to \: \tt \: \lim_{x \to 0} 2 { \bigg(\frac{ \sin x}{x} \bigg)}^{3} \\ \: \\ \to \to \: \tt \: \lim_{x \to 0} 2 \\ \: \\ \bf taking \: limit \: \\ \\ \to \: 2 \\ \\ \to \boxed{ \tt \: \lim_{x \to 0} \frac{ \sin x(1 - \cos2x)}{ {x}^{3} } = 2}$

Another method:-

We have,

$\to \: \tt \: \lim_{x \to 0} \frac{ \sin x(1 - \cos2x)}{ {x}^{3} } \\ \: \\ \sf we \: know \: that \: \\ \\ \to \boxed{ \lim \tt \frac{1 - \cos h}{ {h}^{2} } = \frac{1}{2} } \\ \\ \: \to \: \tt \: \lim_{x \to 0} \: \frac{4}{4} \: \frac{ \sin x(1 - \cos2x)}{ x.{x}^{2} } \\ \: \\ \to \: \tt \: \lim_{x \to 0} 4 \: \frac{ \sin x(1 - \cos2x)}{ 4{x}^{2} .x} \\ \: \\ \to \: \tt \: \lim_{x \to 0} \:4 \: \frac{ \sin x}{ {x} } \times \frac{1}{2} \\ \: \\ \to \: \tt \: \lim_{x \to 0} 2 \: \frac{ \sin x}{ {x}} \\ \: \\ \to \: 2 \\ \\ \to \boxed{ \tt \: \lim_{x \to 0} \frac{ \sin x(1 - \cos2x)}{ {x}^{3} } = 2}$