Question Number 138473 by henderson last updated on 14/Apr/21 $$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{calculate}}\::\:\underset{{x}\rightarrow\infty} {{lim}}\:\left[\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\:\left({x}\right)}\right]^{{x}\:{ln}\:\left({x}\right)} . \\ $$ Answered by mathmax by abdo last updated on 14/Apr/21…
Question Number 138437 by EnterUsername last updated on 13/Apr/21 $$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}\right)!!}=\sqrt{\mathrm{e}} \\ $$ Answered by Ar Brandon last updated on 13/Apr/21 $$\:\:\:\:\:\left(\mathrm{2n}\right)!!=\mathrm{2n}\left(\mathrm{2n}−\mathrm{2}\right)\left(\mathrm{2n}−\mathrm{4}\right)…\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{n}}…
Question Number 72905 by Tony Lin last updated on 04/Nov/19 $${f}\left({x}\right)\geqslant\mathrm{0},\:{and}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0},\underset{{x}\rightarrow{a}} {\mathrm{lim}}{g}\left({x}\right)=\infty \\ $$$${then}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}{f}\left({x}\right)^{{g}\left({x}\right)} =? \\ $$ Commented by mathmax by abdo last…
Question Number 7363 by Yozzia last updated on 25/Aug/16 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{ln}\left({n}\right)}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{tan}^{−\mathrm{1}} {i}}{{n}+\mathrm{1}−{i}}\right)\right\}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 72836 by aliesam last updated on 03/Nov/19 Commented by mathmax by abdo last updated on 04/Nov/19 $${changement}\:{x}=\frac{\pi}{\mathrm{3}}+{t}\:\:{give} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\frac{\sqrt{\mathrm{3}}{cosx}−{sinx}}{\mathrm{1}−\mathrm{2}{cosx}}\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\sqrt{\mathrm{3}}{cos}\left(\frac{\pi}{\mathrm{3}}+{t}\right)−{sin}\left(\frac{\pi}{\mathrm{3}}+{t}\right)}{\mathrm{1}−\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}+{t}\right)} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}}…
Question Number 7200 by Tawakalitu. last updated on 16/Aug/16 $${Prove}\:{that}\: \\ $$$$\frac{{sin}\left(\frac{\Theta}{\mathrm{2}}\right)}{\left(\frac{\Theta}{\mathrm{2}}\right)}\:\:=\:\:\mathrm{1} \\ $$ Commented by Rasheed Soomro last updated on 16/Aug/16 $${Do}\:{you}\:{mean}\:\underset{\Theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\Theta}{\mathrm{2}}\right)\:}{\frac{\Theta}{\mathrm{2}}}=\mathrm{1}\:? \\…
Question Number 72694 by MJS last updated on 31/Oct/19 $$\mathrm{convergent}\:\mathrm{or}\:\mathrm{divergent}? \\ $$$${S}=\frac{\mathrm{2}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{7}}… \\ $$ Commented by mathmax by abdo last updated on 31/Oct/19 $${S}=\sum_{{n}=\mathrm{0}} ^{\infty}…
Question Number 7101 by Tawakalitu. last updated on 10/Aug/16 Commented by Tawakalitu. last updated on 10/Aug/16 $${Working}\:{please}\:………… \\ $$ Answered by Yozzii last updated on…
Question Number 138153 by mathlove last updated on 10/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{x}−\mid{tanx}\mid}{\mid{sinx}\mid−{x}}=? \\ $$ Answered by TheSupreme last updated on 10/Apr/21 $${for}\:{x}\rightarrow\mathrm{0}^{+} \:\mid{sin}\left({x}\right)\mid={sin}\left({x}\right)\:{and}\:\mid{tan}\left({x}\right)\mid={tan}\left({x}\right) \\ $$$${lim}\:\frac{{x}−{tan}\left({x}\right)}{{sin}\left({x}\right)−{x}}=…
Question Number 138140 by liberty last updated on 10/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)−\mathrm{cos}\:{x}}{{x}^{\mathrm{4}} }=? \\ $$ Answered by EDWIN88 last updated on 10/Apr/21 $$\:{recall}\:\mathrm{cos}\:{A}−\mathrm{cos}\:{B}\:=\:\mathrm{2sin}\:\left(\frac{{B}−{A}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{A}+{B}}{\mathrm{2}}\right) \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\left(\frac{{x}−\mathrm{sin}\:{x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{2}}\right)}{{x}^{\mathrm{4}}…