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Category: Limits

1-lim-x-pi-4-sin-pi-4-x-tan-x-pi-4-2-lim-x-pi-2-pi-pi-2x-tan-x-pi-2-2-x-pi-cos-2-x-3-lim-x-0-3x-cos-3x-cos-7x-sin-2x-1-tan-2x-1-4-lim-x-0-sin-2

Question Number 111960 by john santu last updated on 05/Sep/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right).\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\:? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\pi\left(\pi−\mathrm{2}{x}\right)\mathrm{tan}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}\left({x}−\pi\right)\mathrm{cos}\:^{\mathrm{2}} {x}}\:? \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{x}\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{7}{x}\right)}{\:\sqrt{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{1}}\:−\sqrt{\mathrm{tan}\:\mathrm{2}{x}+\mathrm{1}}}\:? \\ $$$$\left(\mathrm{4}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{3}−\sqrt{\mathrm{3}{x}+\mathrm{9}}}\:? \\ $$$$ \\…

bemath-lim-x-0-cos-4x-1-3-1-cos-3x-cos-9x-

Question Number 111923 by bemath last updated on 05/Sep/20 $$\sqrt{{bemath}\:} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4}{x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{9}{x}}\:? \\ $$ Answered by bobhans last updated on 05/Sep/20 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3x}\:−\:\mathrm{cos}\:\mathrm{9x}}\:? \\…

lim-x-0-8x-3-4-sin-6x-tan-6x-4-

Question Number 111914 by bemath last updated on 05/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}+\mathrm{sin}\:\mathrm{6}{x}}\:−\sqrt{\mathrm{tan}\:\mathrm{6}{x}+\mathrm{4}}}\:? \\ $$ Answered by bobhans last updated on 05/Sep/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}+\mathrm{sin}\:\mathrm{6x}}\:−\sqrt{\mathrm{4}+\mathrm{tan}\:\mathrm{6x}}}\:=\: \\…

bemath-1-lim-x-x-2-x-1-x-2-x-3-2-prove-that-n-2-2-n-for-n-N-by-mathematical-induction-

Question Number 111882 by bemath last updated on 05/Sep/20 $$\:\:\sqrt{{bemath}} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{1}}\:−\:\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{3}}\:\right]\:? \\ $$$$\left(\mathrm{2}\right)\:{prove}\:{that}\:{n}^{\mathrm{2}} \:\leqslant\:\mathrm{2}^{{n}} \:{for}\:\forall{n}\in\mathbb{N} \\ $$$${by}\:{mathematical}\:{induction} \\ $$ Answered by…

pls-help-me-L-lim-x-1-x-1-13-x-1-7-x-1-5-x-1-3-

Question Number 46330 by Saorey last updated on 24/Oct/18 $$\mathrm{pls}\:\mathrm{help}\:\mathrm{me}! \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{13}}]{\mathrm{x}}−\sqrt[{\mathrm{7}}]{\mathrm{x}}}{\:\sqrt[{\mathrm{5}}]{\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{x}}} \\ $$ Commented by MJS last updated on 24/Oct/18 $$\mathrm{I}\:\mathrm{would}\:\mathrm{again}\:\mathrm{recommend}\:\mathrm{l}'\mathrm{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{1}}…

n-1-1-n-n-1-n-please-help-me-

Question Number 46187 by annika0209 last updated on 22/Oct/18 $$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}×\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{n}}} }=?\:\:\:\:\:\:\:\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}!!!! \\ $$$$ \\ $$ Commented by maxmathsup by imad last updated on…

Question-46136

Question Number 46136 by Meritguide1234 last updated on 21/Oct/18 Answered by ajfour last updated on 21/Oct/18 $$\left(\mathrm{cos}\:{nx}\right)^{\mathrm{1}/{n}} \:=\:\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \:\frac{{nx}}{\mathrm{2}}\right)^{\mathrm{1}/{n}} \\ $$$$\:\:\:\:\:{if}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:{then}\:\:=\:\mathrm{1}−\frac{\mathrm{2}}{{n}}\left(\frac{{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{4}}\right) \\…