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Question Number 56332 by maxmathsup by imad last updated on 14/Mar/19

calculate lim_(x→0)    ((arctan(1+cosx)−arctan(2+x))/x^3 )

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{arctan}\left(\mathrm{1}+{cosx}\right)−{arctan}\left(\mathrm{2}+{x}\right)}{{x}^{\mathrm{3}} } \\ $$

Answered by kaivan.ahmadi last updated on 07/Apr/19

∼lim_(x→0) ((1+cosx−2−x)/x^3 )=  lim_(x→0) ((−(1−cosx)−x)/x^3 )∼  lim_(x→0) ((−(x^2 /2)−x)/x^3 )=lim_(x→0) ((−x−2)/(2x^2 ))=  ((−2)/0^+ )=−∞

$$\sim{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}+{cosx}−\mathrm{2}−{x}}{{x}^{\mathrm{3}} }= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{−\left(\mathrm{1}−{cosx}\right)−{x}}{{x}^{\mathrm{3}} }\sim \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}}{{x}^{\mathrm{3}} }={lim}_{{x}\rightarrow\mathrm{0}} \frac{−{x}−\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}} }= \\ $$$$\frac{−\mathrm{2}}{\mathrm{0}^{+} }=−\infty \\ $$

Commented by maxmathsup by imad last updated on 07/Apr/19

no sir this limit is real not infinte...

$${no}\:{sir}\:{this}\:{limit}\:{is}\:{real}\:{not}\:{infinte}... \\ $$

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