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Category: Limits

Question-199393

Question Number 199393 by Calculusboy last updated on 02/Nov/23 Answered by cortano12 last updated on 03/Nov/23 $$\:\mathrm{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{2011}}\right)\left(\mathrm{1}+\frac{\mathrm{2x}}{\mathrm{2012}}\right)\left(\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{2013}}\right)−\mathrm{1}}{\mathrm{x}} \\ $$$$\:\:\mathrm{A}=\:\frac{\mathrm{1}}{\mathrm{2011}}\:+\frac{\mathrm{2}}{\mathrm{2012}}\:+\:\frac{\mathrm{3}}{\mathrm{2013}}\: \\ $$ Commented by essaad…

lim-n-0-n-1-x-2-n-n-dx-

Question Number 199084 by universe last updated on 27/Oct/23 $$\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{n}}} \:\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}}\right)^{\mathrm{n}} \mathrm{dx}\:\:\:=\:\:\:\:??? \\ $$ Answered by witcher3 last updated on 27/Oct/23 $$\mathrm{x}=\sqrt{\mathrm{n}}.\mathrm{y}…

lim-x-0-sin-3x-tan-6x-

Question Number 198952 by ArifinTanjung last updated on 26/Oct/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{tan}\:\mathrm{6x}}\:=\:….? \\ $$ Answered by Mathspace last updated on 26/Oct/23 $${lim}=\frac{\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({sin}\left(\mathrm{3}{x}\right)\sim\mathrm{3}{x}\:{and}\:{tan}\left(\mathrm{6}{x}\right)\sim\mathrm{6}{x}\right) \\ $$…

a-n-2-a-n-a-n-1-n-1-n-N-and-here-a-1-and-a-2-then-prove-that-lim-n-a-n-2-2-1-3-

Question Number 198152 by universe last updated on 11/Oct/23 $$\:\:\:\:{a}_{{n}+\mathrm{2}} \:=\:\:\:\sqrt{{a}_{{n}} ×{a}_{{n}+\mathrm{1}} }\:\:\:\forall\:{n}\geqslant\mathrm{1}\:,\:{n}\:\in\:\mathrm{N} \\ $$$$\:\mathrm{and}\:\mathrm{here}\:\:\mathrm{a}_{\mathrm{1}\:} =\:\alpha\:\:{and}\:{a}_{\mathrm{2}} =\:\beta\:\:\mathrm{then} \\ $$$$\:\:\:\mathrm{prove}\:\mathrm{that}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}+\mathrm{2}} \:\:\:=\:\:\left(\alpha×\beta^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$ Answered…

lim-x-1-4025x-1-x-1-x-1-1-x-2-1-4025x-

Question Number 198032 by cortano12 last updated on 08/Oct/23 $$\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4025x}}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{2}}\:}\:+…\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4025x}}}\:\right)=? \\ $$ Commented by universe last updated on 08/Oct/23 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{4025}}}×\frac{\mathrm{1}}{{x}}\underset{{r}=\mathrm{0}} {\overset{\mathrm{4024}{x}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{r}/{x}}} \\…