Question

lim x tends to 0 1-cos2x/x​

Answer 1

$\displaystyle \sf \lim_{x\to 0}\ \dfrac{1-cos\ 2x}{x}$

When we sub. x value directly in this equation , it leads to indeterminant form $\dfrac{0}{0}$ ,

So , Apply L-Hospital's rule ,

$:\implies \displaystyle \sf \lim_{x\to 0} \dfrac{\frac{d}{dx}(1-cos\ 2x)}{\frac{d}{dx}(x)}$

$:\implies \displaystyle \sf \lim_{x\to 0} \dfrac{0-(-2sin\ 2x)}{1}$

$:\implies \displaystyle \sf \lim_{x\to 0} \dfrac{2sin\ 2x}{1}$

$:\implies \sf \dfrac{2sin\ (2(0) )}{1}$

$:\implies \pink{0}\ \; \bigstar$

Answer 2

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\boxed{\tt \:  ⟼ \: (1) \: \: 1 - cosx = 2 {sin}^{2} \dfrac{x}{2} }$

$\boxed{\tt \:  (2) \: \lim_{x\to 0}\dfrac{sinx}{x} = 1}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  ⟼\lim_{x\to 0}\dfrac{1 - cos2x}{x}$

$\large\underline{\bold{Step :- 1 }}$

☆ When we substitute x = 0, we get

$\tt \:  = \dfrac{1 - cos0}{0} = \dfrac{1 - 1}{0} = \dfrac{0}{0}$

$\tt \:  ⟼which \: is \: indeterminant \: form.$

─━─━─━─━─━─━─━─━─━─━─━─━

$\large\underline{\bold{Step :- 2 }}$

$\tt \:  ⟼\lim_{x\to 0}\dfrac{1 - cos2x}{x}$

$\tt \:  = \lim_{x\to 0}\dfrac{2 {sin}^{2} x}{x}$

$\: \: \: \: \: \: \: \: \: \: \: \tt \:  [ \because \: 1 - cosx = 2 {sin}^{2}\dfrac{x}{2} ]$

$\tt \:  = 2 \times \lim_{x\to 0}\bigg( \dfrac{sinx}{x} \bigg) \times \lim_{x\to 0}(sinx)$

$\tt \:  = 2 \times 1 \times 0 = 0$

─━─━─━─━─━─━─━─━─━─━─━─━─