Question Number 35848 by rahul 19 last updated on 24/May/18 $${Find}\:{no}.\:{of}\:{points}\:{of}\:{discontinuity} \\ $$$${of}\:: \\ $$$${f}\left({x}\right)=\:\mathrm{cos}\:\mid{x}\mid\:+\:\mid\mathrm{cos}\:{x}\mid\:+\:\mid\mathrm{cos}\:{x}\mid^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:\mid\mathrm{cos}\:{x}\mid^{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$${in}\:{the}\:{interval}\:\left[−\mathrm{2},\mathrm{3}\right]. \\ $$ Commented by rahul 19 last…
Question Number 101366 by bobhans last updated on 02/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} −\mathrm{sin}\:^{\mathrm{3}} {x}}{{x}^{\mathrm{5}} }\:=? \\ $$ Commented by Dwaipayan Shikari last updated on 02/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 166828 by qaz last updated on 28/Feb/22 $$\mathrm{calculate}:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}+\mathrm{k}^{\alpha} }\:\:\:\:\:\:\:\:\:\:\:\:.\left(\mathrm{0}<\alpha<\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 35736 by rahul 19 last updated on 22/May/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sec}\:\left({ex}\right)\mathrm{sec}\:\left({e}^{\mathrm{2}} {x}\right)……\mathrm{sec}\:\left({e}^{\mathrm{50}} {x}\right)\right)}{{e}^{\mathrm{2}} −{e}^{\mathrm{2cos}\:{x}} }\:=\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 23/May/18…
Question Number 35727 by abdo mathsup 649 cc last updated on 22/May/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{sin}\left({shx}\right)\:−{sh}\left({sinx}\right)}{{x}} \\ $$ Commented by abdo mathsup 649 cc last updated on…
Question Number 35726 by abdo mathsup 649 cc last updated on 22/May/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}−{cos}\left({sinx}\right)}{{x}^{\mathrm{2}} } \\ $$ Commented by abdo mathsup 649 cc last updated…
Question Number 35715 by math1967 last updated on 22/May/18 $${Find}\underset{{n}\rightarrow\infty} {\:{lim}}\frac{\mathrm{3}^{{n}+\mathrm{1}} +\mathrm{2}^{{n}+\mathrm{1}} }{\mathrm{3}^{{n}} +\mathrm{2}^{{n}} } \\ $$ Commented by abdo mathsup 649 cc last updated…
Question Number 101239 by M±th+et+s last updated on 01/Jul/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+……+\frac{\mathrm{1}}{{n}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}……+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}}\right) \\ $$ Answered by mathmax by abdo last updated on 01/Jul/20 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{\mathrm{n}}\:=\mathrm{H}_{\mathrm{n}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+….+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…..+\frac{\mathrm{1}}{\mathrm{2n}}+\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−…−\frac{\mathrm{1}}{\mathrm{2n}}…
Question Number 166771 by cortano1 last updated on 27/Feb/22 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{sin}\:\mathrm{x}}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166735 by qaz last updated on 26/Feb/22 $$\mathrm{For}\:\:\mathrm{some}\:\mathrm{constant}\:\alpha\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\mathrm{calculate}:\:\:\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\frac{\mathrm{t}^{\alpha} }{\alpha}−\mathrm{xt}} \mathrm{dt}\sim\sqrt{\frac{\mathrm{2}\pi}{\mathrm{1}−\alpha}}\centerdot\mathrm{x}^{−\frac{\alpha}{\mathrm{2}\left(\mathrm{1}−\alpha\right)}−\mathrm{1}} \mathrm{e}^{\frac{\mathrm{1}−\alpha}{\alpha}\centerdot\mathrm{x}^{−\frac{\alpha}{\mathrm{1}−\alpha}} } ,\mathrm{x}\rightarrow\mathrm{0}^{+} \\ $$ Terms of Service Privacy…