Menu Close

Category: Limits

lim-x-0-e-2x-4e-x-2x-3-x-3-

Question Number 185537 by mathlove last updated on 23/Jan/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}{x}} −\mathrm{4}{e}^{{x}} +\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{3}} }=? \\ $$ Commented by Ar Brandon last updated on 23/Jan/23 $$\mathrm{Same}\:\mathrm{method}\:\mathrm{as}\:\mathrm{above}…

lim-x-0-4e-3x-9e-2x-6x-5-x-3-

Question Number 185539 by mathlove last updated on 23/Jan/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}{e}^{\mathrm{3}{x}} −\mathrm{9}{e}^{\mathrm{2}{x}} +\mathrm{6}{x}+\mathrm{5}}{{x}^{\mathrm{3}} }=? \\ $$ Answered by Ar Brandon last updated on 23/Jan/23 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}}…

lim-x-x-4-1-5-x-1-1-5-x-1-5-

Question Number 119977 by bramlexs22 last updated on 28/Oct/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt[{\mathrm{5}}]{{x}^{\mathrm{4}} }\left(\sqrt[{\mathrm{5}}]{{x}+\mathrm{1}}−\sqrt[{\mathrm{5}}]{{x}}\:\right)\right)=? \\ $$ Answered by bemath last updated on 28/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{5}}]{{x}^{\mathrm{5}} +{x}^{\mathrm{4}} }−\sqrt[{\mathrm{5}}]{{x}^{\mathrm{5}}…

Question-185510

Question Number 185510 by mathlove last updated on 23/Jan/23 Answered by mahdipoor last updated on 23/Jan/23 $$\mathrm{6}{x}−\mathrm{8}{sin}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)= \\ $$$$\mathrm{6}{x}−\mathrm{8}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+….\right)+ \\ $$$$\left(\mathrm{2}{x}−\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{32}{x}^{\mathrm{5}}…

Question-54341

Question Number 54341 by Meritguide1234 last updated on 02/Feb/19 Answered by iv@0uja last updated on 03/Feb/19 $$\underset{{t}=\mathrm{0}} {\overset{{s}} {\sum}}\:_{{s}} {C}_{{t}} \left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{{t}} =\left(\mathrm{1}+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{{s}} =\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)^{{s}} \\ $$$$\underset{{s}=\mathrm{0}}…