Question Number 185106 by mathlove last updated on 17/Jan/23 Commented by mathlove last updated on 17/Jan/23 $${answer}\:{this}\:{problum}??? \\ $$ Commented by Frix last updated on…
Question Number 119517 by bemath last updated on 25/Oct/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\mathrm{sin}\:\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{sin}\:{x}\:\right]\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({sin}\left({x}+\frac{\mathrm{1}}{{x}}\right)−{sinx}\right) \\ $$$$=\underset{{x}\rightarrow\infty}…
Question Number 119482 by liberty last updated on 24/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}}\right)\:? \\ $$ Answered by Olaf last updated on 25/Oct/20 $$\forall\:{x}\in\mathbb{R}^{\ast} ,\:−\mathrm{1}\:\leqslant\:\mathrm{cos}\left(\frac{\mathrm{1}}{{x}}\right)\:\leqslant\:+\mathrm{1} \\ $$$$\forall\:{x}\in\mathbb{R}^{\ast}…
Question Number 184960 by mathlove last updated on 14/Jan/23 Answered by cortano1 last updated on 15/Jan/23 $${L}=−\frac{\mathrm{9}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{4}−\mathrm{1}\right) \\ $$ Commented by mathlove last updated on…
Question Number 119366 by bobhans last updated on 24/Oct/20 $$\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{{x}+\mathrm{5}}}−\sqrt{\mathrm{3}}}{{x}+\mathrm{1}}\:=? \\ $$ Answered by bobhans last updated on 24/Oct/20 Answered by mathmax by abdo…
Question Number 184901 by mathlove last updated on 13/Jan/23 $$\underset{\frac{\mathrm{2}}{{x}}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{100}+{x}}{\mathrm{101}{x}} \\ $$ Commented by SEKRET last updated on 13/Jan/23 $$\boldsymbol{\mathrm{x}}\rightarrow\mathrm{1} \\ $$ Answered by…
Question Number 119356 by Lordose last updated on 23/Oct/20 Commented by Dwaipayan Shikari last updated on 23/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}{log}\left(\frac{{x}^{{x}} }{{x}!}\right) \\ $$$$=\frac{\mathrm{1}}{{x}}.{log}\left(\frac{{x}^{{x}} {e}^{{x}} }{{x}^{{x}} .\sqrt{\mathrm{2}\pi{x}}}\right)\left({Stirling}'{s}\:{approximation}\:\underset{{n}\rightarrow\infty}…
Question Number 184891 by mathlove last updated on 13/Jan/23 $$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{{x}−\pi}{{sinx}}=? \\ $$ Answered by Mathspace last updated on 13/Jan/23 $${chang}.\:{x}−\pi={t}\:{give} \\ $$$${lim}_{{x}\rightarrow\pi} \frac{{x}−\pi}{{sinx}}={lim}_{{t}\rightarrow\mathrm{0}} \frac{{t}}{{sin}\left(\pi+{t}\right)}…
Question Number 184881 by mathlove last updated on 13/Jan/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}}{sinx}}{{x}}\right)^{\frac{{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{{x}}} =? \\ $$ Answered by aba last updated on 13/Jan/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{2x}}=\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 184883 by mathlove last updated on 13/Jan/23 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt{{x}−\mathrm{2}}}{{x}−\mathrm{2}}=? \\ $$ Answered by aba last updated on 13/Jan/23 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}−\mathrm{2}}}{\mathrm{x}−\mathrm{2}}=\underset{{x}\rightarrow\mathrm{2}^{+} } {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}−\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{0}^{+} }=+\infty…