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 $$\:\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{12}\left(\mathrm{6}{x}−\pi\right)}\:=\:\underset{{x}\rightarrow\pi/\mathrm{6}}…
Question Number 126685 by bramlexs22 last updated on 23/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{{x}\:\mathrm{sin}\:{x}}\:=? \\ $$ Answered by liberty last updated on 23/Dec/20 $$\:\left(\ast\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{{x}\:\mathrm{sin}\:{x}}\:= \\ $$$$\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}}…
Question Number 126599 by liberty last updated on 22/Dec/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left({x}+\:\sqrt{\frac{{x}^{\mathrm{3}} }{{x}−\mathrm{1}}}\:\right) \\ $$ Answered by bramlexs22 last updated on 22/Dec/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}+{x}\sqrt{\frac{{x}}{{x}−\mathrm{1}}}\:\right)\:= \\ $$$$\underset{{x}\rightarrow\infty}…
Question Number 192115 by mnjuly1970 last updated on 08/May/23 $$ \\ $$$$\:\Omega\:=\:\mathrm{lim}_{\:{x}\rightarrow\mathrm{0}} \:\left(\:\:\frac{\:\mathrm{cot}^{\:−\mathrm{1}} \:\left(\frac{\mathrm{1}}{{x}}\:\right)}{\:{x}}\:\right)^{\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }} =\:?\:\:\:\:\:\: \\ $$$$\:\: \\ $$$$ \\ $$ Answered by mehdee42…
Question Number 60981 by necx1 last updated on 28/May/19 Commented by Prithwish sen last updated on 28/May/19 $$\mathrm{Let}\: \\ $$$$\mathrm{A}=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \left(\frac{\mathrm{x}!}{\mathrm{x}^{\mathrm{x}} }\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$\therefore\mathrm{lnA}=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}………..\mathrm{x}}{\mathrm{x}.\mathrm{x}.\mathrm{x}…………\mathrm{x}}\right)…
Question Number 192033 by sonukgindia last updated on 06/May/23 Answered by mehdee42 last updated on 06/May/23 $${L}={e}^{\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{{sinx}}{{x}}\:−\mathrm{1}\:\right)\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:}} \:={e}^{\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{{sinx}\:−\mathrm{1}}{{x}^{\mathrm{3}} }\:\right)\:} =\:{e}^{\:{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }{{x}^{\mathrm{3}}…