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Category: Limits

lim-x-0-sin-x-x-2-

Question Number 133923 by bemath last updated on 25/Feb/21 limx0sinxx2=? Answered by EDWIN88 last updated on 25/Feb/21 $$\:\mathrm{x}^{\mathrm{2}} \:=\:\mid\mathrm{x}\mid^{\mathrm{2}} \:\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mid\mathrm{sin}\:\mathrm{x}\mid}{\mid\mathrm{x}\mid}\:.\frac{\mathrm{1}}{\mid\mathrm{x}\mid}=\:\underset{{x}\rightarrow\mathrm{0}}…

Question-133840

Question Number 133840 by Algoritm last updated on 24/Feb/21 Answered by Dwaipayan Shikari last updated on 24/Feb/21 limx2xxΓ(x+1)4Γ(3x3)3Γ(x+1)6=xx(logx+1)Γ(x+1)+Γ(x+1)xx12Γ(3x3)3Γ(x+1)$$=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)+\mathrm{4}\Gamma'\left(\mathrm{3}\right)−\mathrm{12}\Gamma'\left(\mathrm{3}\right)}{\mathrm{3}\Gamma'\left(\mathrm{3}\right)}=\frac{\mathrm{8}\left({log}\left(\mathrm{2}\right)+\mathrm{1}\right)}{\mathrm{6}\left(\frac{\mathrm{3}}{\mathrm{2}}−\gamma\right)}−\frac{\mathrm{8}}{\mathrm{3}} \