Question Number 130965 by Hilolaxon last updated on 31/Jan/21 Answered by bramlexs22 last updated on 31/Jan/21 $$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{−{x}^{\mathrm{5}} \left(−\mathrm{8}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\frac{\mathrm{4}}{{x}^{\mathrm{5}} }\right)}{−{x}^{\mathrm{5}} \left(−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\frac{\mathrm{7}}{{x}^{\mathrm{5}} }\right)}\:=−\infty\:…
Question Number 65356 by aliesam last updated on 28/Jul/19 Commented by mathmax by abdo last updated on 29/Jul/19 $${let}\:{A}\left({x}\right)\:=\frac{\sqrt{{cos}^{\mathrm{4}} {x}−{cos}\left(\mathrm{2}{x}\right)}}{{ln}\left(\mathrm{1}+\mathrm{3}{x}\right)}\:\:\:\:{we}\:{have}\: \\ $$$${cos}^{\mathrm{4}} {x}\:−{cos}\left(\mathrm{2}{x}\right)\:=\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} −{cos}\left(\mathrm{2}{x}\right) \\…
Question Number 130869 by EDWIN88 last updated on 30/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:{x}}{{x}}\right)^{\frac{\mathrm{tan}\:{x}}{{x}−\mathrm{tan}\:{x}\:}} ? \\ $$ Answered by benjo_mathlover last updated on 30/Jan/21 $$\:\mathrm{ln}\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{x}−\mathrm{tan}\:\mathrm{x}}\right)\mathrm{ln}\:\left(\frac{\mathrm{tan}\:\mathrm{x}}{\mathrm{x}}\right) \\ $$$$\:\mathrm{ln}\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 65324 by hovea cw last updated on 28/Jul/19 Commented by mathmax by abdo last updated on 28/Jul/19 $${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}}…
Question Number 130854 by mathlove last updated on 29/Jan/21 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{{x}} \centerdot{x}!−\mathrm{4}\left(\mathrm{3}{x}−\mathrm{4}\right)!}{\mathrm{3}{x}!−\mathrm{6}} \\ $$ Answered by Dwaipayan Shikari last updated on 29/Jan/21 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{x}^{{x}} \left(\mathrm{1}+{logx}\right){x}!+{x}^{{x}}…
Question Number 130852 by Algoritm last updated on 29/Jan/21 Answered by Dwaipayan Shikari last updated on 29/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{cos}\frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }=\mathrm{2}\left(\frac{{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{4}}}{{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$ Answered…
Question Number 130843 by EDWIN88 last updated on 29/Jan/21 $$\:{Without}\:{L}'{H}\hat {{o}pital}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{7}}\:−\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{{x}−\mathrm{1}}\:?\: \\ $$ Answered by mathmax by abdo last updated…
Question Number 130837 by EDWIN88 last updated on 29/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)−\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{x}^{\mathrm{3}} }\:? \\ $$ Answered by bemath last updated on 29/Jan/21 $$\left(\mathrm{1}\right)\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}}…
Question Number 130696 by LYKA last updated on 28/Jan/21 Answered by mathmax by abdo last updated on 28/Jan/21 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} }\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x},\mathrm{x}\right)=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \frac{\mathrm{x}^{\mathrm{2}}…
Question Number 130679 by LYKA last updated on 28/Jan/21 Commented by liberty last updated on 28/Jan/21 $$\:\underset{{x},\mathrm{y},\mathrm{z}\rightarrow\left(\mathrm{0},\mathrm{1},\mathrm{2}\right)} {\mathrm{lim}e}^{\mathrm{xy}} .\mathrm{cos}\:\left(\frac{\pi\left(\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{3}}\right)\:?\: \\ $$ Commented by…