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Question Number 121052 by mathocean1 last updated on 05/Nov/20
show that  lim_(x→0)  ((1−cosx)/x^2 )=(1/2)
$$\mathrm{show}\:\mathrm{that} \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 05/Nov/20
lim_(x→0) ((1−cosx)/x^2 )=lim_(x→0) ((2sin^2 (x/2))/x^2 )=lim_(x→0) ((2x^2 )/(4x^2 ))=(1/2)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cosx}}{{x}^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by malwan last updated on 05/Nov/20
lim_(x→0)  ((2sin^2 (x/2))/x^2 ) = 2((1/2))^2 = 2×(1/4) = (1/2)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }\:=\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by 675480065 last updated on 05/Nov/20
x→0, cosx→1−(x^2 /2)+(x^4 /(24))+0(x^4 )  ⇒lim_(x→0) ((1−cosx)/x^2 )=lim_(x→0) ((1−1+(x^2 /2)−(x^4 /(24)))/x^2 )=lim_(x→0) ((x^2 ((1/2)−(x^2 /(24))))/x^2 )  =lim_(x→0) ((1/2)−(x^2 /(24)))=(1/2)
$$\mathrm{x}\rightarrow\mathrm{0},\:\mathrm{cosx}\rightarrow\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{24}}+\mathrm{0}\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{24}}}{\mathrm{x}^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{24}}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{24}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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