Question

$\huge\bold \red{{ \displaystyle \lim_{x \to0}}( \frac{1 - \cos2x}{2x \sin2x} )}$




❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​​​​​​​​​​​​​​​​​​​​

Answer 1

[ Kindly see this answer from brainly app ]

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given : \\ \\ \displaystyle \lim_{x \to \: 0} \: \rm \: \left( \frac{1 - cos2x}{2x \: sin2x} \right) \\ \\ = \displaystyle \lim_{x \to \: 0} \rm \: \frac{1}{2} \left( \frac{1 - cos2x}{x \: sin2x} \right) \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: \to \:we \: know \: that : \\ \\ \hookrightarrow \: \sf \: \rm1 - cos2 \theta = 2 {sin}^{2} { \theta} \\ \\ \hookrightarrow \: \sf \: \rm \: sin2 \theta = 2sin \theta.cos \theta\end{array}}}} \\ \pink{ \sf \: apply \: this} \\ \\ = \frac{1}{2}\displaystyle \lim_{x \to \: 0} \rm \: \frac{2 \: {sin}^{2} x}{x.2 \: sinx \: cos \: x} \\ \\ = \frac{1}{2} \displaystyle \lim_{x \to \: 0} \rm \: \frac{sin \: x}{x. \: cos \: \: x} \\ \\ = \frac{1}{2} \: \displaystyle \lim_{x \to \: 0} \rm \: \left ( \: \frac{sin \: x}{x} \times \frac{1}{cos \: x} \right) \\ \\ = \frac{1}{2} \times \displaystyle \lim_{x \to \: 0} \rm \: \frac{sin \: x}{x} \times \displaystyle \lim_{x \to \: 0} \rm \: \frac{1}{cos \: x} \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: \to \:we \: know \: that : \\ \\ \displaystyle \lim_{x \to \: 0} \rm \frac{sin \: x}{x} = 1 \end{array}}}} \\ \pink{ \sf \: apply \: this} \\ \\ = \frac{1}{2} \times 1 \times \displaystyle \lim_{x \to \: 0} \rm \frac{1}{cos \: x} \\ \\ \sf \: apply \: limit \\ \\ = \frac{1}{2} \times \frac{1}{ \rm \: cos \: 0} \\ \\ = \frac{1}{2} \times \frac{1}{1} \\ \\ = \frac{1}{2} \end{array}}}}$

Answer 2

$\bold \red{\lim_{x \to 0} \frac{1 - \cos{\left(2 x \right)}}{2 x \sin{\left(2 x \right)}}}$

$\bold \red{\color{red}{\lim_{x \to 0} \frac{1 - \cos{\left(2 x \right)}}{2 x \sin{\left(2 x \right)}}} = \color{red}{\left(\frac{\lim_{x \to 0} \frac{1 - \cos{\left(2 x \right)}}{x \sin{\left(2 x \right)}}}{2}\right)}}$

$\bold \red{\frac{\color{red}{\lim_{x \to 0} \frac{1 - \cos{\left(2 x \right)}}{x \sin{\left(2 x \right)}}}}{2} = \frac{\color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(1 - \cos{\left(2 x \right)}\right)}{\frac{d}{dx}\left(x \sin{\left(2 x \right)}\right)}}}{2}}$

$\bold \red{\frac{\color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(1 - \cos{\left(2 x \right)}\right)}{\frac{d}{dx}\left(x \sin{\left(2 x \right)}\right)}}}{2} = \frac{\color{red}{\lim_{x \to 0} \frac{2 \sin{\left(2 x \right)}}{2 x \cos{\left(2 x \right)} + \sin{\left(2 x \right)}}}}{2}}$

$\bold \red{\frac{\color{red}{\lim_{x \to 0} \frac{2 \sin{\left(2 x \right)}}{2 x \cos{\left(2 x \right)} + \sin{\left(2 x \right)}}}}{2} = \frac{\color{red}{\left(2 \lim_{x \to 0} \frac{\sin{\left(2 x \right)}}{2 x \cos{\left(2 x \right)} + \sin{\left(2 x \right)}}\right)}}{2}}$

$\color{red}{\lim_{x \to 0} \frac{\sin{\left(2 x \right)}}{2 x \cos{\left(2 x \right)} + \sin{\left(2 x \right)}}} = \color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\sin{\left(2 x \right)}\right)}{\frac{d}{dx}\left(2 x \cos{\left(2 x \right)} + \sin{\left(2 x \right)}\right)}}$

$\color{red}{\lim_{x \to 0} \frac{\frac{d}{dx}\left(\sin{\left(2 x \right)}\right)}{\frac{d}{dx}\left(2 x \cos{\left(2 x \right)} + \sin{\left(2 x \right)}\right)}} = \color{red}{\lim_{x \to 0} \frac{2 \cos{\left(2 x \right)}}{- 4 x \sin{\left(2 x \right)} + 4 \cos{\left(2 x \right)}}}$

$\color{red}{\lim_{x \to 0} \frac{2 \cos{\left(2 x \right)}}{- 4 x \sin{\left(2 x \right)} + 4 \cos{\left(2 x \right)}}} = \color{red}{\lim_{x \to 0}\left(- \frac{\cos{\left(2 x \right)}}{2 x \sin{\left(2 x \right)} - 2 \cos{\left(2 x \right)}}\right)}$

$\color{red}{\lim_{x \to 0}\left(- \frac{\cos{\left(2 x \right)}}{2 x \sin{\left(2 x \right)} - 2 \cos{\left(2 x \right)}}\right)} = \color{red}{\left(\frac{1}{2}\right)}$

$\bold \red{Therefore,}$

$\bold \red{\lim_{x \to 0} \frac{1 - \cos{\left(2 x \right)}}{2 x \sin{\left(2 x \right)}} = \frac{1}{2}}$