Question Number 36563 by bshahid010@gmail.com last updated on 03/Jun/18 Commented by prof Abdo imad last updated on 03/Jun/18 $${let}\:{use}\:{the}\:{changement}\:\:\frac{{a}}{{x}}\:={t} \\ $$$${lim}_{{x}\rightarrow{a}} \left(\mathrm{2}−\frac{{a}}{{x}}\right)^{{tan}\left(\frac{\pi{x}}{\mathrm{2}{a}}\right)} \:={lim}_{{t}\rightarrow\mathrm{1}} \left(\mathrm{2}−{t}\right)^{{tan}\left(\:\frac{\pi}{\mathrm{2}{t}}\right)} \\…
Question Number 167624 by mnjuly1970 last updated on 21/Mar/22 $$ \\ $$$$\:\:{solve} \\ $$$$\:\:\Omega\:={lim}_{\:{n}\rightarrow\infty} {n}^{\:\mathrm{2}} .\:{ln}\left(\:{n}\:.\:{sin}\left(\frac{\mathrm{1}}{{n}}\right)\right)=? \\ $$$$ \\ $$ Answered by qaz last updated…
Question Number 167617 by mathlove last updated on 21/Mar/22 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right]^{{cotx}} =? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{1}}{{x}}\right]=? \\ $$ Answered by cortano1 last updated on 21/Mar/22 $$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 167615 by cortano1 last updated on 20/Mar/22 $$\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}+\mathrm{x}}\:\sqrt[{\mathrm{3}}]{\mathrm{7}+\mathrm{x}^{\mathrm{3}} }−\mathrm{4}}{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:=? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167593 by cortano1 last updated on 20/Mar/22 Answered by qaz last updated on 20/Mar/22 $$\mathrm{A}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{17}−\mathrm{2}\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }\centerdot\sqrt[{\mathrm{3}}]{\mathrm{3}\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{3}} +\mathrm{3}}−\mathrm{9}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{9}−\mathrm{2x}^{\mathrm{2}} −\mathrm{8x}}\centerdot\sqrt[{\mathrm{3}}]{\mathrm{3x}^{\mathrm{3}}…
Question Number 102036 by Study last updated on 06/Jul/20 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\sqrt[{\mathrm{5}}]{{x}^{\mathrm{2}} −\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:+\sqrt{{x}+\mathrm{1}}}=? \\ $$ Answered by john santu last updated on 06/Jul/20 $${L}'{Hopital}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 102034 by Study last updated on 06/Jul/20 $${li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{4}}\:−\mathrm{2}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=? \\ $$ Answered by Dwaipayan Shikari last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{4}}−\mathrm{2}}{\mathrm{2}{x}+\mathrm{4}−\mathrm{8}}.\frac{\mathrm{2}{x}−\mathrm{4}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{8}^{\frac{−\mathrm{2}}{\mathrm{3}}} .\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{18}}…
Question Number 102020 by Dwaipayan Shikari last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered by john santu last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\left(\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)−\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{{x}}} \\…
Question Number 101981 by Rohit@Thakur last updated on 05/Jul/20 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$ Answered by Dwaipayan Shikari last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}…
Question Number 101977 by Ar Brandon last updated on 05/Jul/20 $$\mathrm{Given}\:\mathrm{2}\:\mathrm{functions},\:\mathrm{f}\:\mathrm{and}\:\mathrm{g},\:\mathrm{n}-\mathrm{times}\:\mathrm{derivable}\:\mathrm{within} \\ $$$$\mathrm{the}\:\mathrm{open}\:\mathrm{interval},\:\mathbb{R}\:\mathrm{and}\:\mathrm{verify}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{f}^{\left(\mathrm{k}\right)} \left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{0}\:,\:\mathrm{g}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{g}^{\left(\mathrm{k}\right)} \left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{0}\:,\:\forall\mathrm{k}\in\left\{\mathrm{1},\mathrm{2},…,\mathrm{n}−\mathrm{1}\right\} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\underset{\mathrm{x}\rightarrow\mathrm{x}_{\mathrm{0}} } {\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\mathrm{f}^{\left(\mathrm{n}\right)}…