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Question Number 58671 by Mikael_Marshall last updated on 27/Apr/19

lim_(x→0) ((1−cos5x)/x^2 )

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\:\frac{\mathrm{1}−{cos}\mathrm{5}{x}}{{x}^{\mathrm{2}} } \\ $$

Commented by maxmathsup by imad last updated on 27/Apr/19

we have 1−cos(5x) ∼ (((5x)^2 )/2) (x ∈V(0)) ⇒lim_(x→0) ((1−cos(5x))/x^2 ) =((25)/2)

$${we}\:{have}\:\mathrm{1}−{cos}\left(\mathrm{5}{x}\right)\:\sim\:\frac{\left(\mathrm{5}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:\:\left({x}\:\in{V}\left(\mathrm{0}\right)\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}−{cos}\left(\mathrm{5}{x}\right)}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{25}}{\mathrm{2}} \\ $$

Commented by Mikael_Marshall last updated on 28/Apr/19

thanks Sir

$${thanks}\:{Sir} \\ $$

Answered by mr W last updated on 27/Apr/19

=lim_(x→0) ((5 sin 5x)/(2x)) =lim_(x→0) ((sin 5x)/(5x))×((25)/2) =((25)/2)

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{5}\:\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{2}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{5}{x}}×\frac{\mathrm{25}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{25}}{\mathrm{2}} \\ $$

Commented by Mikael_Marshall last updated on 27/Apr/19

thanks Sir

$${thanks}\:{Sir} \\ $$

Answered by ajfour last updated on 27/Apr/19

=((25)/4)lim_(x→0) ((2sin^2 (((5x)/2)))/((((5x)/2))^2 )) =((25)/4)×2=((25)/2) .

$$=\frac{\mathrm{25}}{\mathrm{4}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{5x}}{\mathrm{2}}\right)}{\left(\frac{\mathrm{5x}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\frac{\mathrm{25}}{\mathrm{4}}×\mathrm{2}=\frac{\mathrm{25}}{\mathrm{2}}\:. \\ $$

Commented by Mikael_Marshall last updated on 27/Apr/19

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

Answered by malwaan last updated on 28/Apr/19

lim_(x→0) ((1−cos5x)/x^2 )×((1+cos5x)/(1+cos5x)) lim_(x→0) ((sin^2 5x)/(x^2 (1+cos5x))) =(5^2 /(1+1)) =((25)/2)

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{1}−{cos}\mathrm{5}{x}}{{x}^{\mathrm{2}} }×\frac{\mathrm{1}+{cos}\mathrm{5}{x}}{\mathrm{1}+{cos}\mathrm{5}{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}^{\mathrm{2}} \mathrm{5}{x}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{cos}\mathrm{5}{x}\right)} \\ $$$$=\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{1}+\mathrm{1}}\:=\frac{\mathrm{25}}{\mathrm{2}} \\ $$

Commented by Mikael_Marshall last updated on 28/Apr/19

thankz Sir

$${thankz}\:{Sir} \\ $$

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