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Question Number 58016 by rahul 19 last updated on 16/Apr/19

Value of lim_(x→0) ((cosh x−cos x)/(xsin x)) =?

$${Value}\:{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cosh}\:{x}−\mathrm{cos}\:{x}}{{x}\mathrm{sin}\:{x}}\:=? \\ $$

Commented by rahul 19 last updated on 16/Apr/19

by L−H rule , i′m getting 0. kindly check...

$$\:{by}\:{L}−{H}\:{rule}\:,\:{i}'{m}\:{getting}\:\mathrm{0}. \\ $$$${kindly}\:{check}... \\ $$

Answered by mr W last updated on 16/Apr/19

lim_(x→0) ((sinh x+sin x)/(sin x+x cos x))=((lim_(x→0) ((sinh x)/(sin x))+1)/(1+lim_(x→0) (x/(sin x))×cos x))=((1+1)/(1+1))=1

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sinh}\:{x}+\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}+{x}\:\mathrm{cos}\:{x}}=\frac{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sinh}\:{x}}{\mathrm{sin}\:{x}}+\mathrm{1}}{\mathrm{1}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{\mathrm{sin}\:{x}}×\mathrm{cos}\:{x}}=\frac{\mathrm{1}+\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\mathrm{1} \\ $$

Commented by rahul 19 last updated on 16/Apr/19

thanks sir.

$${thanks}\:{sir}. \\ $$

Answered by tanmay last updated on 16/Apr/19

lim_(x→0) ((((e^x +e^(−x) )/2)−cosx)/(xsinx)) (1/2)lim_(x→0) ((e^x +e^(−x) −2cosx)/(xsinx)) (1/2)lim_(x→0) (((1+x+(x^2 /(2!))+(x^3 /(3!))+...)+(1−x+(x^2 /(2!))−(x^3 /(3!))+..)−2(1−(x^2 /(2!))+(x^4 /(4!))−..))/(x(x−(x^3 /(3!))+(x^5 /(5!))−...))) (1/2)lim_(x→0) (((2+((2x^2 )/(2!))+((2x^4 )/(4!))+...)−(2−((2x^2 )/(2!))+((2x^4 )/(4!))+...))/(x^2 (1−(x^2 /(3!))+(x^4 /(5!))+..))) (1/2)lim_(x→0) ((2x^2 +((2x^6 )/(6!))+...)/(x^2 (1−(x^2 /(3!))+(x^4 /(5!))))) =(1/2)×((2+0)/((1−0)))=1 or method or (1/2)lim_(x→0) ((e^x +e^(−x) −2cosx)/(xsinx))((0/0)) =(1/2)lim_(x→0) ((e^x −e^(−x) +2sinx)/(xcosx+sinx)) =(1/2)lim_(x→0) (((e^x −1)−(e^(−x) −1)+2sinx)/(xcosx+sinx)) (1/2)lim_(x→0) ((((e^x −1)/x)+((e^(−x) −1)/(−x))+2((sinx)/x))/(cosx+((sinx)/x))) (1/2)×((1+1+2×1)/(1+1)) (1/2)×((4/2))=1 (1/2)lim_(x→0)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}−{cosx}}{{xsinx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} +{e}^{−{x}} −\mathrm{2}{cosx}}{{xsinx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+...\right)+\left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+..\right)−\mathrm{2}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}−..\right)}{{x}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−...\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{4}!}+...\right)−\left(\mathrm{2}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{4}!}+...\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}+..\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{2}{x}^{\mathrm{6}} }{\mathrm{6}!}+...}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}+\mathrm{0}}{\left(\mathrm{1}−\mathrm{0}\right)}=\mathrm{1} \\ $$$${or}\:{method} \\ $$$${or}\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} +{e}^{−{x}} −\mathrm{2}{cosx}}{{xsinx}}\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −{e}^{−{x}} +\mathrm{2}{sinx}}{{xcosx}+{sinx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({e}^{{x}} −\mathrm{1}\right)−\left({e}^{−{x}} −\mathrm{1}\right)+\mathrm{2}{sinx}}{{xcosx}+{sinx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{e}^{{x}} −\mathrm{1}}{{x}}+\frac{{e}^{−{x}} −\mathrm{1}}{−{x}}+\mathrm{2}\frac{{sinx}}{{x}}}{{cosx}+\frac{{sinx}}{{x}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}+\mathrm{1}+\mathrm{2}×\mathrm{1}}{\mathrm{1}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{4}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}} \\ $$

Commented by rahul 19 last updated on 16/Apr/19

Thank U sir.

$${Thank}\:{U}\:{sir}. \\ $$

Commented by tanmay last updated on 16/Apr/19

most welcome

$${most}\:{welcome}\: \\ $$

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