Question Number 118950 by benjo_mathlover last updated on 21/Oct/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } .{e}^{−{x}} \:=\:?. \\ $$ Answered by john santu last updated on 21/Oct/20 $$\:\mathcal{L}\:=\:\underset{{x}\rightarrow\infty}…
Question Number 53376 by rajeshghorai130@gmail.com last updated on 21/Jan/19 Commented by Abdo msup. last updated on 21/Jan/19 $${let}\:{A}\left({x}\right)=\frac{\left({sin}\alpha\right)^{{x}} −\left({cos}\alpha\right)^{{x}} \:−{cos}\left(\mathrm{2}\alpha\right)}{{x}−\mathrm{4}} \\ $$$${A}\left({x}\right)=_{{x}−\mathrm{4}={t}} \:\:\:\:\frac{\left({sin}\alpha\right)^{{t}+\mathrm{4}} \:−\left({cos}\alpha\right)^{{t}+\mathrm{4}} −{cos}\left(\mathrm{2}\alpha\right)}{{t}}={B}\left({t}\right)…
Question Number 184435 by moh777 last updated on 06/Jan/23 Answered by CElcedricjunior last updated on 06/Jan/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{arsinx}}−\boldsymbol{{arctanx}}}{\boldsymbol{{x}}}=\mathrm{0} \\ $$$$ \\ $$ Terms of Service…
Question Number 184383 by universe last updated on 05/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 184329 by greougoury555 last updated on 05/Jan/23 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left({p}+{x}\right)−\mathrm{tan}^{−\mathrm{1}} \left({p}−{x}\right)}{{x}}=? \\ $$ Answered by SEKRET last updated on 05/Jan/23 $$\:\:\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{p}}^{\mathrm{2}} } \\…
Question Number 184239 by mathlove last updated on 04/Jan/23 $${f}\left({x}\right)=\begin{cases}{{x}−\mathrm{2}\:\:\:\:\:\:\:{if}\:\:{x}>\mathrm{3}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:{x}=\mathrm{3}}\\{−{x}+\mathrm{4}\:\:\:\:{if}\:\:\mathrm{1}\leqslant{x}<\mathrm{3}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:\mathrm{1}<{x}}\end{cases} \\ $$$${faind}\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=?\:\:{and}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}{f}\left({x}\right)=? \\ $$ Answered by SEKRET last updated on 04/Jan/23 $$\:\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{1}^{−} }…
Question Number 118681 by bemath last updated on 19/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:−\:\sqrt{\mathrm{sin}\:{x}}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:=? \\ $$ Answered by benjo_mathlover last updated on 19/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:×\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:+\sqrt{\mathrm{sin}\:{x}}}\:= \\…
Question Number 118427 by bramlexs22 last updated on 17/Oct/20 $$\:\:\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{{x}^{\mathrm{2}} \:}]{\mathrm{cos}\:{x}}\: \\ $$$$\:\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{cos}\:\pi{x}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 183936 by universe last updated on 31/Dec/22 Answered by Ar Brandon last updated on 31/Dec/22 $$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\overset{{n}} {\:}{C}_{{k}} }\right)^{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}}…
Question Number 183927 by greougoury555 last updated on 31/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{2}\:} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{12}}\:=? \\ $$ Answered by cortano1 last updated on 31/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{6}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}}…