Question Number 68528 by Joel122 last updated on 13/Sep/19 $$\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{{t}}\:\int_{\mathrm{1}} ^{\:{t}} \:\sqrt[{{x}}]{{t}}\:{dx}\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 133997 by rs4089 last updated on 26/Feb/21 $${lim}_{{n}\rightarrow\infty} \left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{n}}}{{n}^{\mathrm{2}} }\right)^{{n}} \\ $$ Answered by mathmax by abdo last updated on 27/Feb/21 $$\mathrm{U}_{\mathrm{n}} =\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+….+\frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{n}^{\mathrm{2}}…
Question Number 2888 by prakash jain last updated on 29/Nov/15 $$\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}=? \\ $$$$\mathrm{or} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\Re\left({z}\right)}{\mid{z}\mid} \\ $$ Answered by prakash jain…
Question Number 133923 by bemath last updated on 25/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mid\mathrm{sin}\:\mathrm{x}\mid}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$ Answered by EDWIN88 last updated on 25/Feb/21 $$\:\mathrm{x}^{\mathrm{2}} \:=\:\mid\mathrm{x}\mid^{\mathrm{2}} \:\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mid\mathrm{sin}\:\mathrm{x}\mid}{\mid\mathrm{x}\mid}\:.\frac{\mathrm{1}}{\mid\mathrm{x}\mid}=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 68354 by TawaTawa last updated on 09/Sep/19 $$\underset{{x},\mathrm{y}\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{4}} \:−\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{4}} \mathrm{y}^{\mathrm{4}} \:+\:\mathrm{y}^{\mathrm{2}} } \\ $$ Commented by kaivan.ahmadi last…
Question Number 2806 by prakash jain last updated on 27/Nov/15 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{cos}\:{ix}}{{i}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{uniformly}\:\mathrm{convergent}\:\mathrm{on}\:\mathrm{real}\:\mathrm{line}. \\ $$ Answered by Yozzi last updated on 27/Nov/15…
Question Number 2783 by 123456 last updated on 27/Nov/15 $$\mathrm{call}\:\gamma:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}H}_{{n}} −\mathrm{ln}\:{n} \\ $$$$\mathrm{proof}\:\mathrm{that}\:\gamma\:\mathrm{is}\:\mathrm{finite}\:\mathrm{and}\:\gamma\in\left(\mathrm{0},\mathrm{1}\right) \\ $$ Commented by Filup last updated on 27/Nov/15 $$\mathrm{I}\:\mathrm{am}\:\mathrm{curious}\:\mathrm{as}\:\mathrm{to}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these} \\…
Question Number 133840 by Algoritm last updated on 24/Feb/21 Answered by Dwaipayan Shikari last updated on 24/Feb/21 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{{x}} \Gamma\left({x}+\mathrm{1}\right)−\mathrm{4}\Gamma\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}=\frac{{x}^{{x}} \left({logx}+\mathrm{1}\right)\Gamma\left({x}+\mathrm{1}\right)+\Gamma'\left({x}+\mathrm{1}\right){x}^{{x}} −\mathrm{12}\Gamma'\left(\mathrm{3}{x}−\mathrm{3}\right)}{\mathrm{3}\Gamma'\left({x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)+\mathrm{4}\Gamma'\left(\mathrm{3}\right)−\mathrm{12}\Gamma'\left(\mathrm{3}\right)}{\mathrm{3}\Gamma'\left(\mathrm{3}\right)}=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)}{\mathrm{6}\left(\frac{\mathrm{3}}{\mathrm{2}}−\gamma\right)}−\frac{\mathrm{8}}{\mathrm{3}} \\…
Question Number 133830 by metamorfose last updated on 24/Feb/21 $${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−{cos}\left({x}\right){cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right){cos}^{\mathrm{3}} \left(\mathrm{3}{x}\right)…{cos}^{{n}} \left({nx}\right)}{{x}^{\mathrm{2}} }=…? \\ $$ Answered by EDWIN88 last updated on 24/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 2735 by prakash jain last updated on 25/Nov/15 $$\mathrm{Does}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{converge}? \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\mid\frac{\mathrm{sin}\:{i}}{{i}}\mid \\ $$ Commented by Filup last updated on 26/Nov/15 $$\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{limit}…