Question

lim x tends to 0 x²/1-cosx​

Answer 1

GIVEN :–

$\\\implies \displaystyle \bf \lim_{x \to0} \bigg[ \dfrac{ {x}^{2} }{1 - \cos(x) } \bigg]$

TO FIND :–

• Value of limit = ?

SOLUTION :–

• Let –

$\\\implies \displaystyle \bf P = \lim_{x \to0} \bigg[ \dfrac{ {x}^{2} }{1 - \cos(x) } \bigg]$

• After putting limits it's in $\dfrac{0}{0}$ , Hence Applying L'HOSPITAL Rule –

$\\\implies \displaystyle \bf P = \lim_{x \to0} \bigg[ \dfrac{2x}{0 - \{ - \sin(x) \}} \bigg]$

$\\\implies \displaystyle \bf P = \lim_{x \to0} \bigg[ \dfrac{2x}{0 + \sin(x)} \bigg]$

$\\\implies \displaystyle \bf P = \lim_{x \to0} \bigg[ \dfrac{2x}{\sin(x)} \bigg]$

• After putting limits it's in $\dfrac{0}{0}$ , Hence Again Applying L'HOSPITAL Rule –

$\\\implies \displaystyle \bf P = \lim_{x \to0} \bigg[ \dfrac{2}{\cos(x)} \bigg]$

• Now put limits –

$\\\implies\bf P =\dfrac{2}{\cos(0)}$

• We know that –

$\\\large \longrightarrow {\boxed{ \bf\cos(0) =1}}$

• So that –

$\\\implies\bf P =\dfrac{2}{1}$

$\\\implies \large{ \boxed{\bf P =2}}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{\lim _{x\:\rightarrow \:0}\left[\dfrac{x^2}{1-\cos \left(x\right)}\right]}}$

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{\lim _{x\to \:0}\left(\frac{x^2}{1-\cos \left(x\right)}\right)=2}}$

★═════════════════★

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\lim _{x\to \:0}\left(\dfrac{x^2}{1-\cos \left(x\right)}\right)}$

$\text{Apply L'Hopital's Rule}\\\\ =\lim _{x \rightarrow 0}\left(\dfrac{2 x}{\sin (x)}\right) \\\\\\\text{Apply L'Hopital's Rule}\\\\ =\lim _{x \rightarrow 0}\left(\dfrac{2}{\cos (x)}\right)$

$\mathrm{Use\:the\:following\:identity}:\quad \dfrac{1}{\cos \left(x\right)}=\sec \left(x\right)$

$=\lim _{x\to \:0}\left(2\sec \left(x\right)\right)$

$\mathrm{Plug\:in\:the\:value}\:x=0$

$=2\sec \left(0\right)$

$\star \sec (0)=1\star$

$=2\cdot \:1$

$\huge\boxed{\sf{=2}}$