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Category: Limits

calculate-L-lim-n-1-1-2-1-1-3-1-1-n-1-n-

Question Number 197947 by mnjuly1970 last updated on 05/Oct/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\ldots \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \sqrt[{{n}}]{\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\ldots\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:=\:?\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$ Answered by MM42 last updated on…

lim-x-2-cosx-1-cosx-

Question Number 197832 by mathlove last updated on 30/Sep/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}+\sqrt{{cosx}}}{−\mathrm{1}+\sqrt{{cosx}}}=? \\ $$ Answered by MM42 last updated on 30/Sep/23 $${let}\:\:{f}\left({x}\right)=\frac{\mathrm{2}+\sqrt{{cosx}}}{−\mathrm{1}+\sqrt{{cosx}}} \\ $$$${for}\:\:{a}_{{n}} =\mathrm{2}{n}\pi\Rightarrow{lim}_{{n}\rightarrow\infty} \:{f}\left({a}_{{n}}…

lim-n-0-1-dt-1-t-t-n-

Question Number 197661 by pticantor last updated on 25/Sep/23 $$\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}+….+{t}^{{n}} }=?\: \\ $$ Commented by JDamian last updated on 25/Sep/23 $$\mathrm{Why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{post}\:\mathrm{again}\:\mathrm{the}\:\mathrm{same}\:\mathrm{topic} \\…

Question-197660

Question Number 197660 by pticantor last updated on 25/Sep/23 Answered by witcher3 last updated on 25/Sep/23 $$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}+…\mathrm{t}^{\mathrm{n}} }\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{dt}=\mathrm{1} \\ $$$$\underset{\mathrm{n}\rightarrow\infty}…

lim-x-4x-4x-2-5-2x-1-bx-

Question Number 197548 by cortano12 last updated on 21/Sep/23 $$\:\:\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{4x}−\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{5}}}{\mathrm{2x}−\mathrm{1}}\right)^{\mathrm{bx}} =?\: \\ $$ Answered by MM42 last updated on 21/Sep/23 $${lim}_{{x}\rightarrow−\infty} \:\left(\frac{\mathrm{4}{x}−\mid\mathrm{2}{x}\mid}{\mathrm{2}{x}−\mathrm{1}}\right)^{{bx}} ={lim}_{{x}\rightarrow−\infty}…