Question Number 58239 by salahahmed last updated on 20/Apr/19 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left({x}^{\frac{\mathrm{1}}{{x}}} \right) \\ $$ Answered by salahahmed last updated on 23/Apr/19 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left({e}^{\frac{\mathrm{ln}\left({x}\right)}{{x}}}…
Question Number 189285 by TUN last updated on 14/Mar/23 $${f}\left({x}\right)\:{is}\:{continous}\:{function}\:{on}\:{R} \\ $$$${and}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{f}\left(\frac{{x}+\mathrm{1}}{{x}}\right)−\mathrm{6}}{\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$${Evalute}\::\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt{{f}\left({x}\right)+{x}}−{x}}{\left({x}−\mathrm{1}\right)}=¿ \\ $$ Answered by cortano12 last updated on…
Question Number 189145 by mathlove last updated on 12/Mar/23 $${pleas}\:{solve}\:{this} \\ $$$$\left.\mathrm{1}\right)\:\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{e}^{{x}+\mathrm{2}{x}+\mathrm{3}{x}+\mathrm{4}{x}+\centerdot\centerdot\centerdot\centerdot\centerdot+{nx}} −{e}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} }{{x}−\mathrm{1}}=? \\ $$$$\left.\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} \centerdot\mathrm{4}^{{x}} \centerdot\centerdot\centerdot\centerdot{n}^{{x}} } −{e}^{{n}!} }{{x}−\mathrm{1}}=? \\…
Question Number 58067 by mustakim420 last updated on 17/Apr/19 Commented by maxmathsup by imad last updated on 17/Apr/19 $$\left.{let}\:{A}\left({x}\right)\:={x}^{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} −{e}} \:\:\:\Rightarrow{A}\left({x}\right)\:=\:{e}^{\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} −{e}\right\}{ln}\left({x}\right)} \:\:\Rightarrow{ln}\left({A}\left({x}\right)\right)=\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} −{e}\right)\right\}{ln}\left({x}\right) \\…
Question Number 58016 by rahul 19 last updated on 16/Apr/19 $${Value}\:{of}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cosh}\:{x}−\mathrm{cos}\:{x}}{{x}\mathrm{sin}\:{x}}\:=? \\ $$ Commented by rahul 19 last updated on 16/Apr/19 $$\:{by}\:{L}−{H}\:{rule}\:,\:{i}'{m}\:{getting}\:\mathrm{0}. \\ $$$${kindly}\:{check}……
Question Number 57957 by rahul 19 last updated on 15/Apr/19 Commented by Smail last updated on 15/Apr/19 $${L}_{{n}} =\frac{\mathrm{2}^{{n}} +\left(−\mathrm{2}\right)^{{n}} }{\mathrm{3}^{{n}} }=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right) \\…
Question Number 188985 by normans last updated on 10/Mar/23 $$ \\ $$$$\:\:\:\boldsymbol{\mathrm{Lim}}_{\boldsymbol{{x}}\rightarrow\sim} \:\:\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{4}}{\mathrm{9}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}}\:=\:??\: \\ $$ Commented by MJS_new last updated on…
Question Number 188984 by normans last updated on 10/Mar/23 $$ \\ $$$$\:\:\boldsymbol{\mathrm{Lim}}_{\boldsymbol{{x}}\rightarrow\sim} \:\sqrt{\mathrm{16}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}−\mathrm{1}}−\mathrm{4}\boldsymbol{{x}}−\mathrm{5}\:=\:??\:\:\:\: \\ $$$$ \\ $$ Commented by MJS_new last updated on 13/Mar/23…
Question Number 57866 by Mikael_Marshall last updated on 13/Apr/19 $$\underset{{n}\rightarrow\infty} {{lim}}\:\:\frac{\mathrm{3}^{{n}+\mathrm{2}} −\mathrm{2}.\mathrm{5}^{{n}+\mathrm{1}} }{\mathrm{3}^{{n}} −\mathrm{2}.\mathrm{5}^{{n}−\mathrm{1}} } \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19 $$\underset{{n}\rightarrow\infty}…
Question Number 123398 by bemath last updated on 25/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{3}{x}}\:−\sqrt{\mathrm{1}+\mathrm{2}{x}}}{{x}^{\mathrm{2}} }\:? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Nov/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{3}}+\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}!}.\left(−\frac{\mathrm{2}}{\mathrm{9}}\right)−\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{2}}+\frac{\mathrm{4}{x}^{\mathrm{2}}…