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if-f-is-continuos-and-diferentiable-everywhere-on-R-if-f-0-0-and-f-x-f-x-then-proof-that-f-x-0-

Question Number 532 by 123456 last updated on 25/Jan/15 $${if}\:{f}\:{is}\:{continuos}\:{and}\:{diferentiable} \\ $$$${everywhere}\:{on}\:\mathbb{R},\:{if}\:{f}\left(\mathrm{0}\right)=\mathrm{0}\:{and} \\ $$$$\mid{f}'\left({x}\right)\mid\leqslant\mid{f}\left({x}\right)\mid\:{then}\:{proof}\:{that} \\ $$$${f}\left({x}\right)=\mathrm{0} \\ $$ Answered by prakash jain last updated on…

Prove-or-disprove-n-0-1-n-2-97-2-2-97-e-97-e-97-2-388-e-2-97-1-e-2-97-1-37635-37636-1-388-97-

Question Number 131580 by Dwaipayan Shikari last updated on 06/Feb/21 $$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{97}\right)^{\mathrm{2}} }=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{97}\left(\boldsymbol{{e}}^{\boldsymbol{\pi}\sqrt{\mathrm{97}}} −{e}^{−\boldsymbol{\pi}\sqrt{\mathrm{97}}} \right)^{\mathrm{2}} }+\frac{\boldsymbol{\pi}}{\mathrm{388}}.\frac{{e}^{\mathrm{2}\boldsymbol{\pi}\sqrt{\mathrm{97}}} +\mathrm{1}}{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{\pi}\sqrt{\mathrm{97}}} −\mathrm{1}}+\frac{\mathrm{37635}}{\mathrm{37636}}−\frac{\mathrm{1}}{\:\mathrm{388}\sqrt{\mathrm{97}}} \\ $$…

lim-x-1-2-x-x-

Question Number 66019 by Rio Michael last updated on 07/Aug/19 $$\underset{{x}\rightarrow\infty} {{lim}}\:\left(\mathrm{1}\:+\:\frac{\mathrm{2}}{{x}}\right)^{{x}} \:= \\ $$ Commented by mathmax by abdo last updated on 07/Aug/19 $${let}\:{f}\left({x}\right)=\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)^{{x}}…

proof-or-given-a-counter-example-if-x-n-is-a-no-limited-sequence-then-exist-a-sub-sequence-x-nk-that-lim-n-0-1-x-nk-0-

Question Number 481 by 123456 last updated on 12/Jan/15 $${proof}\:{or}\:{given}\:{a}\:{counter}\:{example}: \\ $$$${if}\:\left\{{x}_{{n}} \right\}\:{is}\:{a}\:{no}\:{limited}\:{sequence} \\ $$$${then} \\ $$$${exist}\:{a}\:{sub}−{sequence}\:\left\{{x}_{{nk}} \right\}\:{that} \\ $$$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}_{{nk}} }=\mathrm{0} \\ $$ Commented…

show-that-the-equation-xe-x-1-has-a-root-between-0-5-and-0-6-starting-with-0-55-as-a-first-approximate-

Question Number 66017 by Rio Michael last updated on 07/Aug/19 $${show}\:{that}\:{the}\:{equation}\:{xe}^{{x}} =\mathrm{1}\:{has}\:{a}\:{root}\:{between}\:\mathrm{0}.\mathrm{5}\:{and}\:\mathrm{0}.\mathrm{6}\:{starting} \\ $$$${with}\:\mathrm{0}.\mathrm{55}\:{as}\:{a}\:{first}\:{approximate}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

given-a-n-and-b-n-two-real-sequence-can-a-serie-n-1-a-n-and-n-1-b-n-diverge-but-n-1-a-n-b-n-converge-

Question Number 476 by 123456 last updated on 11/Jan/15 $${given}\:{a}_{{n}} \:{and}\:{b}_{{n}} \:{two}\:{real}\:{sequence} \\ $$$${can}\:{a}\:{serie}\:\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}{a}_{{n}} \:{and}\:\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}{b}_{{n}} \:{diverge} \\ $$$${but} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty}…