Question Number 197110 by cortano12 last updated on 08/Sep/23 Answered by MathematicalUser2357 last updated on 10/Sep/23 $$\mathrm{let}\:{f}\left({x}\right)=\frac{\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{2}\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{4}}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Then}\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\infty…
Question Number 197054 by cortano12 last updated on 07/Sep/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{9}} \mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:^{\mathrm{7}} \mathrm{x}}{\mathrm{4x}−\pi}\:=? \\ $$ Answered by universe last updated on 07/Sep/23 $$\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}} \\ $$…
Question Number 197008 by tri26112004 last updated on 06/Sep/23 Answered by AST last updated on 06/Sep/23 $$\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{2}}…
Question Number 197029 by cortano12 last updated on 06/Sep/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{8}} −\mathrm{sin}\:^{\mathrm{8}} \mathrm{x}}{\mathrm{x}^{\mathrm{10}} }\:=? \\ $$ Answered by MM42 last updated on 06/Sep/23 $${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\left({x}^{\mathrm{4}}…
Question Number 196959 by cortano12 last updated on 05/Sep/23 $$\:\:\:\:\:\:\underline{ } \\ $$ Answered by horsebrand11 last updated on 05/Sep/23 $$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{2x}−\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\:\right)+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{ax}+\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\right)=\mathrm{1}−\mathrm{b}…
Question Number 196806 by cortano12 last updated on 01/Sep/23 $$\:\:\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$ Answered by dimentri last updated on 01/Sep/23 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{1}/\mathrm{6}} \left(\mathrm{2}−\mathrm{2}{x}\right)^{\mathrm{1}/\mathrm{6}} −{x}}{\mathrm{2}{x}\:−{x}\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}+\:\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{3}}…
Question Number 196534 by cortano12 last updated on 27/Aug/23 $$\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\:\sqrt{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{x}}\right)}\:=? \\ $$ Answered by Mathspace last updated on 27/Aug/23 $$\mathrm{1}−{cosu}\sim\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\left({u}\rightarrow{o}\right)\:\Rightarrow \\ $$$$\mathrm{1}−{cos}\left(\frac{\pi}{{x}}\right)\sim\frac{\pi^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}}…
Question Number 196491 by universe last updated on 26/Aug/23 Commented by hardmath last updated on 26/Aug/23 $$=\:\mathrm{2}^{\frac{\mathrm{2}}{\boldsymbol{\pi}}} \: \\ $$ Commented by universe last updated…
Question Number 196471 by Tawa11 last updated on 25/Aug/23 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}}\left(\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{n}}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{2}}{\mathrm{n}}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{3}}{\mathrm{n}}}\:\:+\:\:..\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}\:\:+\:\:\frac{\mathrm{n}}{\mathrm{n}}}\right) \\ $$ Answered by mr W last updated on 25/Aug/23 $$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{2}+{x}} \\…
Question Number 196399 by tri26112004 last updated on 24/Aug/23 $${a}/\:\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{2}\right)} {\mathrm{lim}}\:\left(\mathrm{1}+{xy}\right)^{\frac{\mathrm{2}}{{x}^{\mathrm{2}} +{xy}}} \\ $$$${b}/\:\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){sin}\left(\frac{\mathrm{1}}{{xy}}\right) \\ $$$${c}/\underset{\left({x},{y}\right)\rightarrow\left(\infty,\infty\right)} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){e}^{−\left({x}+{y}\right)} \\ $$ Answered…