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Category: Limits

Question-130965

Question Number 130965 by Hilolaxon last updated on 31/Jan/21 Answered by bramlexs22 last updated on 31/Jan/21 $$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{−{x}^{\mathrm{5}} \left(−\mathrm{8}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\frac{\mathrm{4}}{{x}^{\mathrm{5}} }\right)}{−{x}^{\mathrm{5}} \left(−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\frac{\mathrm{7}}{{x}^{\mathrm{5}} }\right)}\:=−\infty\:…

Question-65356

Question Number 65356 by aliesam last updated on 28/Jul/19 Commented by mathmax by abdo last updated on 29/Jul/19 letA(x)=cos4xcos(2x)ln(1+3x)wehave$${cos}^{\mathrm{4}} {x}\:−{cos}\left(\mathrm{2}{x}\right)\:=\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} −{cos}\left(\mathrm{2}{x}\right) \

Question-65324

Question Number 65324 by hovea cw last updated on 28/Jul/19 Commented by mathmax by abdo last updated on 28/Jul/19 $${we}\:{have}\:\mathrm{1}\leqslant{k}\leqslant{n}^{\mathrm{2}} \:\Rightarrow\mathrm{1}+{n}^{\mathrm{2}} \leqslant{k}+{n}^{\mathrm{2}} \leqslant\mathrm{2}{n}^{\mathrm{2}} \:\Rightarrow\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\leqslant\sqrt{{k}+{n}^{\mathrm{2}}…