Question Number 197407 by mathlove last updated on 16/Sep/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=? \\ $$ Answered by Rasheed.Sindhi last updated on 16/Sep/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\mathrm{1}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 197301 by uchihayahia last updated on 13/Sep/23 $$ \\ $$$$\:{how}\:{do}\:{i}\:{calculate}\:{this} \\ $$$$\:\underset{{x}\rightarrow-\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{2}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{1}} \\ $$$$\:{multiplying}\:{both}\:{numerator} \\ $$$$\:{and}\:{denumerator}\:{by}\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} } \\…
Question Number 197281 by cortano12 last updated on 12/Sep/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{x}+\mathrm{2x}^{\mathrm{5}} }{\mathrm{3x}^{\mathrm{3}} }\:=? \\ $$ Answered by MM42 last updated on 12/Sep/23 $${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{5}}…
Question Number 197282 by cortano12 last updated on 12/Sep/23 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sin}\:\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \:\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:\mathrm{x}^{\mathrm{2}} \:\right)}\:=? \\ $$ Answered by MM42 last updated on 12/Sep/23…
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}}…