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Category: Limits

prove-that-n-1-cos-n-n-1-1-2-1-3-1-n-is-convergent-

Question Number 216274 by mnjuly1970 last updated on 02/Feb/25 $$ \\ $$$$\:\:\:{prove}\:{that}\:: \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{cos}\left(\:{n}\:\right)}{{n}}\:\left(\:\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\:…+\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\right) \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:{is}\:\:\:{convergent}. \\ $$$$ \\…

lim-x-0-sin-2-2x-cos-x-1-3-cos-x-1-4-

Question Number 216055 by efronzo1 last updated on 26/Jan/25 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{x}}}\:=? \\ $$ Answered by golsendro last updated on 27/Jan/25 $$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}−\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{x}}}…

lim-x-0-1-cos-x-cos-2x-x-2-

Question Number 216032 by efronzo1 last updated on 26/Jan/25 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\:\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$ Answered by mr W last updated on 26/Jan/25 $$\mathrm{cos}\:{x}\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\…

lim-x-cos-pi-2-sin-3-x-x-sin-3-x-2-1-x-

Question Number 216014 by mathlove last updated on 25/Jan/25 $$\underset{\Delta{x}\rightarrow{cos}\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{sin}^{\mathrm{3}} \left(\Delta{x}+{x}\right)−{sin}^{\mathrm{3}} {x}}{\mathrm{2}^{−\mathrm{1}} \centerdot\Delta{x}}=? \\ $$ Answered by MrGaster last updated on 26/Jan/25 $$=\mathrm{2}\centerdot\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{3}}…

lim-x-0-cos-2x-cos-6x-1-cos-3x-cos-5x-

Question Number 215884 by golsendro last updated on 20/Jan/25 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{6x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3x}\:\mathrm{cos}\:\mathrm{5x}}\:=? \\ $$$$\:\:\:\: \\ $$ Answered by oubiji last updated on 20/Jan/25 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{6x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{3x}\:\mathrm{cos}\:\mathrm{5x}}\: \\…

lim-x-x-1-3-x-1-3-x-

Question Number 215831 by golsendro last updated on 19/Jan/25 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }−\sqrt{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{3}} }}{\:\sqrt{\mathrm{x}}}\:=? \\ $$ Answered by efronzo1 last updated on 19/Jan/25 $$\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}}…

Question-215808

Question Number 215808 by efronzo1 last updated on 19/Jan/25 $$\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by golsendro last updated on 19/Jan/25 $$\:\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{a}^{{x}} +{b}^{{x}} +{c}^{{x}} −\mathrm{3}}{{x}}\right)} \\…

Question-215779

Question Number 215779 by ucup last updated on 18/Jan/25 Commented by MathematicalUser2357 last updated on 18/Jan/25 The "Sehingga" looks like "Sh-nigga", so I flagged your post as inappropriate. Commented by AntonCWX last updated on 18/Jan/25 $${Sehingga}\:{means}\:{until}\:{in}\:{English}.…

lim-n-k-1-n-1-1-k-1-n-1-k-1-n-1-k-1-n-

Question Number 215748 by MrGaster last updated on 17/Jan/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\prod}}\Gamma\left(\frac{\mathrm{1}}{{k}}\right)\right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} −\left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\Gamma\left(\frac{\mathrm{1}}{{k}}\right)\right]^{\frac{\mathrm{1}}{{n}}} \\ $$ Answered by MrGaster last updated on 02/Feb/25…

Question-215739

Question Number 215739 by universe last updated on 16/Jan/25 Answered by MrGaster last updated on 19/Jan/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}+\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{n}^{{k}} }{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{{n}} } \\…