Question Number 131052 by benjo_mathlover last updated on 01/Feb/21 $$\:\underset{{x}\rightarrow{s}} {\mathrm{lim}}\:\frac{{x}^{{s}^{{s}} } −{s}^{{x}^{{s}} } }{{x}^{\mathrm{2}} −{s}^{\mathrm{2}} }\:=? \\ $$ Answered by Ar Brandon last updated…
Question Number 131033 by greg_ed last updated on 31/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{tan}\:{x}}{{x}\:\mathrm{tan}^{\mathrm{2}} {x}}\:=\:? \\ $$ Commented by greg_ed last updated on 02/Feb/21 $$\boldsymbol{\mathrm{without}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{limited}}\:\boldsymbol{\mathrm{development}}\:\boldsymbol{\mathrm{method}}\:! \\ $$ Answered…
Question Number 65490 by Masumsiddiqui399@gmail.com last updated on 30/Jul/19 Commented by mathmax by abdo last updated on 31/Jul/19 $${hospital}\:{method}\:\:{let}\:{u}\left({x}\right)\:={ln}\left({x}\right)\:{and}\:{v}\left({x}\right)={cos}\left(\frac{\pi}{\mathrm{2}^{{x}} }\right) \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{1}} {u}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{1}} {v}\left({x}\right)\:=\mathrm{0} \\…
Question Number 65472 by Masumsiddiqui399@gmail.com last updated on 30/Jul/19 $$ \\ $$$$ \\ $$$$\:\:\:{solve}\:\: \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left[\left(\mathrm{2}+{x}\right)^{{n}} −\mathrm{2}^{{n}} \right]}{{x}} \\ $$$$ \\ $$ Commented by…
Question Number 65473 by Tawa1 last updated on 30/Jul/19 Commented by mathmax by abdo last updated on 30/Jul/19 $${let}\:\:{A}\left({x}\right)\:=\sqrt{\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}+\mathrm{1}}}−^{\mathrm{3}} \sqrt{\frac{{x}^{\mathrm{5}} +\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}}}\:\Rightarrow{for}\:{x}>\mathrm{0} \\ $$$${A}\left({x}\right)\:=\sqrt{\frac{{x}^{\mathrm{3}}…
Question Number 65464 by aliesam last updated on 30/Jul/19 Commented by mathmax by abdo last updated on 30/Jul/19 $${let}\:\:{A}\left({x}\right)\:={x}\left({x}+\mathrm{1}\right){ln}\left(\frac{{x}+\mathrm{1}}{{x}}\right)−{x}\:\:\:\:{we}\:{have}\:{for}\:{x}\in{V}\left(+\infty\right) \\ $$$${ln}\left(\frac{{x}+\mathrm{1}}{{x}}\right)\:={ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\:=\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+{o}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\: \\ $$$${A}\left({x}\right)\:=\left({x}^{\mathrm{2}}…
Question Number 130965 by Hilolaxon last updated on 31/Jan/21 Answered by bramlexs22 last updated on 31/Jan/21 $$\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}}…