Question Number 11395 by Joel576 last updated on 23/Mar/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{those}\:\mathrm{functions}\:\mathrm{below}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{limit} \\ $$$$\left.\mathrm{a}\right)\:\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\:\frac{{xy}}{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} } \\ $$$$ \\ $$$$\left.{b}\right)\:\:\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\:\frac{{xy}\:+\:{y}^{\mathrm{3}} }{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} } \\ $$…
Question Number 142464 by som(math1967) last updated on 01/Jun/21 $$\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{lim}}}\:\left[\frac{\left(\mathrm{2}\boldsymbol{{n}}\right)!}{\boldsymbol{{n}}!\boldsymbol{{n}}^{\boldsymbol{{n}}} }\right]^{\frac{\mathrm{1}}{\boldsymbol{{n}}}} =? \\ $$ Answered by Dwaipayan Shikari last updated on 01/Jun/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{2}^{\mathrm{2}{n}}…
Question Number 142462 by BHOOPENDRA last updated on 01/Jun/21 Answered by MJS_new last updated on 01/Jun/21 $${y}={px}\:\Rightarrow\:\frac{{xy}+\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{{p}}{{p}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}}{{p}^{\mathrm{2}} +\mathrm{1}}×\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${p}\in\mathbb{R}\:\Rightarrow\:\frac{{p}}{{p}^{\mathrm{2}} +\mathrm{1}}={q}\in\mathbb{R}\wedge{r}=\frac{\mathrm{2}}{{p}^{\mathrm{2}}…
Question Number 142459 by bramlexs22 last updated on 01/Jun/21 $$\:\:\:\:\:\:−−−−−−−−−−− \\ $$$$\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{{x}}{\mathrm{cot}\:{x}}−\frac{\pi}{\mathrm{2cos}\:{x}}\right)=? \\ $$$$\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$ Answered by benjo_mathlover last updated on 01/Jun/21 $$\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}}…
Question Number 11364 by Joel576 last updated on 22/Mar/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\sqrt{\mathrm{1}\:+\:\mathrm{tan}\:{x}}\:−\:\sqrt{\mathrm{1}\:+\:\mathrm{sin}\:{x}}}{{x}^{\mathrm{3}} } \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Mar/17 $$\frac{\sqrt{\mathrm{1}+{tgx}}−\sqrt{\mathrm{1}+{sinx}}}{{x}^{\mathrm{3}} }=\frac{\left(\mathrm{1}+{tgx}\right)−\left(\mathrm{1}+{sinx}\right)}{\left(\sqrt{\mathrm{1}+{tgx}}+\sqrt{\mathrm{1}+{sinx}}\right).{x}^{\mathrm{3}} }= \\…
Question Number 76815 by Rio Michael last updated on 30/Dec/19 $$\mathrm{Which}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{is}\:\boldsymbol{\mathrm{N}}\mathrm{o}\boldsymbol{\mathrm{t}}\:\mathrm{convergent}? \\ $$$$\mathrm{A}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{3}} } \\ $$$$\mathrm{B}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} } \\ $$$$\mathrm{C}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}}…
Question Number 142351 by bramlexs22 last updated on 30/May/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{27}+{x}}−\sqrt[{\mathrm{3}\:}]{\mathrm{27}−{x}}}{\:\sqrt[{\mathrm{3}\:}]{{x}^{\mathrm{2}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{3}} }}\:=? \\ $$ Answered by mathmax by abdo last updated on 30/May/21 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{27}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 142329 by HarshSahu last updated on 30/May/21 $$\:{lim}_{{x}\rightarrow\infty} \:\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered by Dwaipayan Shikari last updated on 30/May/21 $$\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} ={y}…
Question Number 142309 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{x}^{\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{x}} } −\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{x}^{\mathrm{sin}\:\mathrm{x}} } }{\mathrm{x}^{\mathrm{3}} }=? \\ $$ Answered by mnjuly1970 last updated on…
Question Number 142308 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{lnlnln}\left[\mathrm{x}+\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{x}}} \right]+\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{e}+\mathrm{1}} }\right)}{\mathrm{x}^{\mathrm{2}} }=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com