Question

please solve this. ​

Answer 1

Given

$: \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{1 - cosx}{2 {x}^{2} }$

Now Check the form , So put the value x = 0

$: \sf \implies \dfrac{1 - cos0}{2(0) {}^{2} }$

$: \implies \dfrac{1 - 1}{0} = \dfrac{0}{0}$

So it is 0/0 Form , use L'hospital rule

In L'hospital rule we differentiate Numerator and Denominator with Respect to x

$: \implies \displaystyle \sf \lim _{ \sf \: x \to0} \dfrac{ \dfrac{d(1 - cosx)}{dx} }{ \dfrac{d(2 {x}^{2} )}{dx} }$

$: \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{ \dfrac{d(1)}{dx} - \dfrac{d(cosx)}{dx} }{ 2\dfrac{d( {x}^{2} )}{dx} }$

$: \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{0 - ( - sinx)}{2(2x)}$

$: \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{sinx}{4x}$

We know that

$: \implies \displaystyle \sf \lim _{ \sf \: x \to0} \dfrac{sinx^{ } }{x} = 1$

By using this We get

$\sf : \implies 1 \times \dfrac{1}{4} = \dfrac{1}{4}$

Answer

$: \implies \dfrac{1}{4}$

Answer

Answer 2

Question :-

$\red \maltese \: \: \displaystyle \sf \lim _{ \sf \: x \to0} \frac{1 - cosx}{2 {x}^{2} }$

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Given :-

$\red \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{1 - cosx}{2 {x}^{2} }$

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Solution :-

• Check the form,

• So put the value x = 0,

$\red \sf\red \implies \dfrac{1 - cos0}{2(0) {}^{2} }$

$\red \implies \dfrac{1 - 1}{0} = \dfrac{0}{0}$

• So it is 0/0 Form , (using L'hospital rule),
• So it is 0/0 Form , (using L'hospital rule),In L'hospital rule we differentiate Numerator and Denominator with Respect to 'x'

$\red \implies \displaystyle \sf \lim _{ \sf \: x \to0} \dfrac{ \dfrac{d(1 - cosx)}{dx} }{ \dfrac{d(2 {x}^{2} )}{dx} }$

$\red \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{ \dfrac{d(1)}{dx} - \dfrac{d(cosx)}{dx} }{ 2\dfrac{d( {x}^{2} )}{dx} }$

$\red \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{0 - ( - sinx)}{2(2x)}$

$\red \implies \displaystyle \sf \lim _{ \sf \: x \to0} \frac{sinx}{4x}$

• We know that,

$\red \implies \displaystyle \sf \lim _{ \sf \: x \to0} \dfrac{sinx^{ } }{x} = 1$

• By using this We get,

$\sf \red \implies 1 \times \dfrac{1}{4} = \dfrac{1}{4}$

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Final Answer :-

$\dfrac{1}{4}$

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