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Question Number 128607 by john_santu last updated on 08/Jan/21

lim_(x→0) ((x.cot x−1)/x^2 )=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}.\mathrm{cot}\:\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=? \\ $$

Answered by malwan last updated on 08/Jan/21

lim_(x→0)  (((x/(tan x)) − 1)/x^2 ) = lim_(x→0)  ((x−tan x)/(x^2  tan x))  = lim_(x→0)  ((x−(x +(x^3 /3)  +.....))/(x^2  tan x))  = lim_(x→0)   ((−(x/3))/(tan x))  = − (1/3)

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{x}}{{tan}\:{x}}\:−\:\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}−{tan}\:{x}}{{x}^{\mathrm{2}} \:{tan}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}−\left({x}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\:+.....\right)}{{x}^{\mathrm{2}} \:{tan}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\:\frac{−\frac{{x}}{\mathrm{3}}}{{tan}\:{x}}\:\:=\:−\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by liberty last updated on 08/Jan/21

 lim_(x→0) ((xcos x−sin x)/(x^2 sin x))=lim_(x→0) ((x(1−(x^2 /2))−(x−(x^3 /6)))/x^3 )  = −(1/3)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{xcos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\mathrm{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\mathrm{x}^{\mathrm{3}} } \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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