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Category: Limits

Question-120268

Question Number 120268 by bramlexs22 last updated on 30/Oct/20 Answered by mathmax by abdo last updated on 30/Oct/20 fcontinueatx=0f(0)=limx0f(x)=limx0e1+x2cos(1x)$$\mathrm{we}\:\mathrm{have}\:−\mathrm{1}\leqslant\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\leqslant\mathrm{1}\:\Rightarrow−\mathrm{x}^{\mathrm{2}} \leqslant\mathrm{x}^{\mathrm{2}}…

Question-185776

Question Number 185776 by TUN last updated on 27/Jan/23 Answered by MJS_new last updated on 27/Jan/23 n!nnen2πn$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}!{n}^{{n}} }\right)^{\mathrm{1}/{n}} \:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4}×\mathrm{2}^{\mathrm{1}/\left(\mathrm{2}{n}\right)}…

f-x-2-x-3x-x-examine-its-continuity-at-x-1-2-where-x-is-greatest-integer-function-

Question Number 54688 by pooja24 last updated on 09/Feb/19 f(x)=2[x]3x[x]examineitscontinuityatx=12where[x]isgreatestintegerfunction Commented by maxmathsup by imad last updated on 10/Feb/19 $${we}\:{have}\:{f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\mathrm{2}\left(−\mathrm{1}\right)}{−\frac{\mathrm{3}}{\mathrm{2}}−\left(−\mathrm{1}\right)}\:=\frac{−\mathrm{2}}{−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}\:=\frac{−\mathrm{2}}{−\frac{\mathrm{1}}{\mathrm{2}}}\:=\mathrm{4} \