Question Number 58016 by rahul 19 last updated on 16/Apr/19
$${Value}\:{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cosh}\:{x}−\mathrm{cos}\:{x}}{{x}\mathrm{sin}\:{x}}\:=? \\ $$
Commented by rahul 19 last updated on 16/Apr/19
$$\:{by}\:{L}−{H}\:{rule}\:,\:{i}'{m}\:{getting}\:\mathrm{0}. \\ $$$${kindly}\:{check}… \\ $$
Answered by mr W last updated on 16/Apr/19
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sinh}\:{x}+\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}+{x}\:\mathrm{cos}\:{x}}=\frac{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sinh}\:{x}}{\mathrm{sin}\:{x}}+\mathrm{1}}{\mathrm{1}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}}{\mathrm{sin}\:{x}}×\mathrm{cos}\:{x}}=\frac{\mathrm{1}+\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\mathrm{1} \\ $$
Commented by rahul 19 last updated on 16/Apr/19
$${thanks}\:{sir}. \\ $$
Answered by tanmay last updated on 16/Apr/19
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}−{cosx}}{{xsinx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} +{e}^{−{x}} −\mathrm{2}{cosx}}{{xsinx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…\right)+\left(\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+..\right)−\mathrm{2}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}−..\right)}{{x}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−…\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{4}!}+…\right)−\left(\mathrm{2}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\mathrm{4}!}+…\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}+..\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{2}{x}^{\mathrm{6}} }{\mathrm{6}!}+…}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}+\mathrm{0}}{\left(\mathrm{1}−\mathrm{0}\right)}=\mathrm{1} \\ $$$${or}\:{method} \\ $$$${or}\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} +{e}^{−{x}} −\mathrm{2}{cosx}}{{xsinx}}\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −{e}^{−{x}} +\mathrm{2}{sinx}}{{xcosx}+{sinx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({e}^{{x}} −\mathrm{1}\right)−\left({e}^{−{x}} −\mathrm{1}\right)+\mathrm{2}{sinx}}{{xcosx}+{sinx}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{{e}^{{x}} −\mathrm{1}}{{x}}+\frac{{e}^{−{x}} −\mathrm{1}}{−{x}}+\mathrm{2}\frac{{sinx}}{{x}}}{{cosx}+\frac{{sinx}}{{x}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}+\mathrm{1}+\mathrm{2}×\mathrm{1}}{\mathrm{1}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{4}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}} \\ $$
Commented by rahul 19 last updated on 16/Apr/19
$${Thank}\:{U}\:{sir}. \\ $$
Commented by tanmay last updated on 16/Apr/19
$${most}\:{welcome}\: \\ $$