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Category: Limits

prove-that-lim-x-1-sin-cos-x-x-1-2-3-2-

Question Number 78012 by malwaan last updated on 13/Jan/20 $$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{lim}}}\frac{\boldsymbol{{sin}}\left(\boldsymbol{\pi{cos}\pi{x}}\right)}{\left(\boldsymbol{{x}}−\mathrm{1}\right)^{\mathrm{2}} }\:=\:−\:\frac{\boldsymbol{\pi}^{\mathrm{3}} }{\mathrm{2}} \\ $$ Commented by msup trace by abdo last updated…

lim-x-0-1-1-x-2-cos-x-tan-4-x-

Question Number 143532 by bramlexs22 last updated on 15/Jun/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mathrm{cos}\:{x}}{\mathrm{tan}\:^{\mathrm{4}} {x}}\:=? \\ $$ Answered by mathmax by abdo last updated on 15/Jun/21 $$\mathrm{cosx}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}}…

lim-x-2-x-3-x-1-3-5-x-1-5-3x-2-2x-3-1-3-

Question Number 143531 by bramlexs22 last updated on 15/Jun/21 $$\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{{x}}+\mathrm{3}\:\sqrt[{\mathrm{3}}]{{x}}\:+\mathrm{5}\:\sqrt[{\mathrm{5}}]{{x}}}{\:\sqrt{\mathrm{3}{x}−\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{3}}}\:=? \\ $$ Answered by bobhans last updated on 15/Jun/21 Terms of Service Privacy Policy…

Question-143453

Question Number 143453 by yahy last updated on 14/Jun/21 Answered by Willson last updated on 14/Jun/21 $$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}^{\:\:\:\:\mathrm{n}} }\sqrt{\frac{\mathrm{n}^{\mathrm{n}} }{\mathrm{n}!}}\:=\:\frac{\mathrm{1}}{{e}} \\ $$ Commented by yahy…

lim-x-0-ln-cos-3x-ln-cos-2x-

Question Number 12382 by tawa last updated on 21/Apr/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\mathrm{ln}\:\mathrm{cos}\left(\mathrm{3x}\right)}{\mathrm{ln}\:\mathrm{cos}\left(\mathrm{2x}\right)} \\ $$ Answered by ajfour last updated on 21/Apr/17 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left[\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}\:\mathrm{3}{x}\right)\right]}{\mathrm{ln}\:\left[\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right)\right]} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left[\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}}…