Question Number 63190 by Tawa1 last updated on 30/Jun/19 $$\mathrm{Test}\:\mathrm{its}\:\mathrm{convergence}:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} \:\mathrm{sin}^{\mathrm{2}} \mathrm{n}} \\ $$ Commented by mathmax by abdo last updated on 01/Jul/19…
Question Number 128674 by bemath last updated on 09/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4x}−\mathrm{12tan}\:\mathrm{x}}{\mathrm{3sin}\:\mathrm{4x}−\mathrm{12sin}\:\mathrm{x}}\:=\:? \\ $$$$ \\ $$ Answered by liberty last updated on 09/Jan/21 $$\:\mathrm{Taylor}\:\mathrm{series}\:\begin{cases}{\mathrm{tan}\:\mathrm{x}=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2x}^{\mathrm{5}} }{\mathrm{15}}+…}\\{\mathrm{tan}\:\mathrm{4x}=\mathrm{4x}+\frac{\mathrm{64x}^{\mathrm{3}}…
Question Number 128654 by john_santu last updated on 09/Jan/21 $$\:\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\:\:\left[\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{tan}\:\mathrm{x}}\:−\mathrm{1}}{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}}\:\right]\:=? \\ $$ Answered by bemath last updated on 09/Jan/21 Terms of Service Privacy…
Question Number 128626 by john_santu last updated on 09/Jan/21 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left\{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \right\}=? \\ $$$$\left\{\:\right\}\:=\:\mathrm{fractional}\: \\ $$$$ \\ $$ Commented by liberty last updated on 09/Jan/21…
Question Number 128624 by john_santu last updated on 09/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mid\mathrm{x}^{\mathrm{2}} +\mathrm{1}\mid−\mid\mathrm{x}^{\mathrm{2}} −\mathrm{1}\mid}{\mid\mathrm{2000x}+\mathrm{5}\mid−\mid\mathrm{2000x}−\mathrm{5}\mid}\:=? \\ $$ Answered by liberty last updated on 09/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\left(\mathrm{x}^{\mathrm{2}}…
Question Number 128607 by john_santu last updated on 08/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}.\mathrm{cot}\:\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=? \\ $$ Answered by malwan last updated on 08/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{x}}{{tan}\:{x}}\:−\:\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}−{tan}\:{x}}{{x}^{\mathrm{2}}…
Question Number 128605 by john_santu last updated on 08/Jan/21 $$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} }=? \\ $$ Commented by john_santu last updated on 08/Jan/21 Answered by mathmax…
Question Number 128604 by john_santu last updated on 08/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{x}} −\mathrm{x}}{\mathrm{ln}\:\mathrm{x}−\mathrm{x}+\mathrm{1}}\:? \\ $$ Answered by liberty last updated on 09/Jan/21 $$\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{x}} \right)=\mathrm{x}^{\mathrm{x}} \left(\mathrm{ln}\:\mathrm{x}+\mathrm{1}\right) \\…
Question Number 128593 by Eric002 last updated on 08/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mid\mathrm{3}{x}−\mathrm{1}\mid−\mid\mathrm{3}{x}+\mathrm{1}\mid}{{x}} \\ $$ Commented by MJS_new last updated on 08/Jan/21 $$\frac{\mid\mathrm{3}{x}−\mathrm{1}\mid−\mid\mathrm{3}{x}+\mathrm{1}\mid}{{x}}=\frac{\mathrm{3}}{{x}}\left(\mid{x}−\frac{\mathrm{1}}{\mathrm{3}}\mid−\mid{x}+\frac{\mathrm{1}}{\mathrm{3}}\mid\right)= \\ $$$$\:\:\:\:\:\left[−\frac{\mathrm{1}}{\mathrm{3}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{3}}\right] \\ $$$$=\frac{\mathrm{3}}{{x}}\left(−\mathrm{2}{x}\right)=−\frac{\mathrm{6}{x}}{{x}}=−\mathrm{6}\forall{x}\in\left[−\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{1}}{\mathrm{3}}\right]\wedge{x}\neq\mathrm{0}…
Question Number 63054 by jimful last updated on 28/Jun/19 $${if}\:\Sigma\mid{a}_{{n}} \:\mid\:{is}\:{convergent},\:{then} \\ $$$${prove}\:{that}\:{there}\:{exists}\: \\ $$$${a}\:{subsequence}\:\left\{{n}_{{k}} {a}_{{n}_{{k}} } \right\}\:\:{with} \\ $$$$\underset{{k}\rightarrow\infty} {\mathrm{lim}}{n}_{{k}} {a}_{{n}_{{k}} } =\mathrm{0} \\…