Question Number 127009 by BHOOPENDRA last updated on 26/Dec/20 $${f}\left({x},{y}\right)=\left\{\frac{\mathrm{2}{x}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\:\:\:,\left({x},{y}\right)\neq\left(\mathrm{0},\mathrm{0}\right)\right. \\ $$$$\left\{\mathrm{0}\:\:\:\:,\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right)\right. \\ $$$${check}\:{continuity} \\ $$ Answered by Olaf last updated…
Question Number 127002 by bramlexs22 last updated on 26/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}+\mathrm{cos}\:\left(\mathrm{3}{x}\right)−\mathrm{3csch}\:\left({x}\right)}{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=? \\ $$ Commented by Evimene last updated on 26/Dec/20 $$ \\ $$$$\:\: \\…
Question Number 126923 by bramlexs22 last updated on 25/Dec/20 $$\:\:\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\frac{{x}−\mathrm{2cos}\:\sqrt{\mathrm{2}−{x}}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by bemath last updated on 25/Dec/20 Terms of Service…
Question Number 192433 by cortano12 last updated on 18/May/23 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{ln}\:\left(\mathrm{3x}+\mathrm{1}\right)\right)}{\mathrm{3x}^{\mathrm{6}} −\mathrm{tan}\:\left(\mathrm{3x}^{\mathrm{2}} \right)}=? \\ $$ Answered by mehdee42 last updated on 18/May/23 $${tip}\::{if}\:\:{u}\rightarrow\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−{cosu}\sim\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}}…
Question Number 61363 by bhanukumarb2@gmail.com last updated on 01/Jun/19 Answered by tanmay last updated on 02/Jun/19 $${N}_{{r}} =\mathrm{2}^{\mathrm{2}{b}_{{n}} } −\mathrm{2}^{\mathrm{1}+{b}_{{n}} } .\mathrm{3}^{{b}_{{n}} } +\mathrm{3}^{\mathrm{2}{b}_{{n}} }…
Question Number 126824 by MathSh last updated on 24/Dec/20 $$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{f}\left({x}+\mathrm{2}\right)−\mathrm{6}}{{x}−\mathrm{2}}=\mathrm{4}\:\:{if},\:\:\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\mathrm{4}\centerdot{f}\left({x}\right)−\mathrm{24}}{{x}−\mathrm{4}}=? \\ $$ Answered by mahdipoor last updated on 24/Dec/20 $$\underset{{x}\rightarrow\mathrm{2}} {{lim}}\frac{{f}\left({x}+\mathrm{2}\right)−\mathrm{6}}{{x}−\mathrm{2}}=\frac{{f}\left(\mathrm{4}\right)−\mathrm{6}}{\mathrm{0}}=\mathrm{4}\Rightarrow{f}\left(\mathrm{4}\right)−\mathrm{6}=\mathrm{0} \\ $$$$\overset{{L}'{hopital}\:}…
Question Number 126801 by john_santu last updated on 24/Dec/20 $$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\:\frac{{x}^{{a}} −{a}^{{x}} }{{a}^{{x}} −{a}^{{a}} }\:=\:\:;\:{a}>\mathrm{0}\: \\ $$$$\: \\ $$ Answered by liberty last updated on…
Question Number 126761 by bemath last updated on 24/Dec/20 Answered by liberty last updated on 24/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{2}} +{O}\left({x}^{\mathrm{3}} \right)\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{O}\left({x}^{\mathrm{5}} \right)\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} \left({x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}}…
Question Number 192281 by uchihayahia last updated on 14/May/23 $$ \\ $$$$ \\ $$$$\:{help}\:{me}\:{proving}\:{this} \\ $$$$\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:{dan}\:{c}\in\mathbb{R} \\ $$$$\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:{f}\left({x}\right)={L}\:\Leftrightarrow\:\underset{{c}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}+{c}\right)={L} \\ $$$$\: \\ $$$$ \\…
Question Number 126744 by bemath last updated on 24/Dec/20 $$\:\:{Given}\:{f}\left({x}\right)=\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\left(\mathrm{6}{x}−\pi\right)^{\mathrm{2}} } \\ $$$$\:{Find}\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:{f}\left({x}\right)\:. \\ $$ Answered by liberty last updated on 24/Dec/20 $$\:\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{12}\left(\mathrm{6}{x}−\pi\right)}\:=\:\underset{{x}\rightarrow\pi/\mathrm{6}}…