Question Number 137902 by benjo_mathlover last updated on 08/Apr/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:=? \\ $$ Answered by greg_ed last updated on 08/Apr/21 $$ \\ $$$$\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}} \\…
Question Number 137820 by bramlexs22 last updated on 07/Apr/21 $$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }\:? \\ $$ Answered by SLVR last updated on 07/Apr/21 Commented by SLVR last…
Question Number 72251 by aliesam last updated on 26/Oct/19 Commented by mathmax by abdo last updated on 26/Oct/19 $${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{ax}+{bx}^{\mathrm{2}} \right)−{ln}\left(\mathrm{1}−{ax}+{bx}^{\mathrm{2}} \right)}{\mathrm{1}−{cosx}} \\ $$$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}+{u}\right)=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\mathrm{1}−{u}\:+{o}\left({u}\right)\:\left({u}\rightarrow\mathrm{0}\right)\:\Rightarrow \\…
Question Number 137773 by Ajadiazeemolamilekan last updated on 06/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{px}−\mathrm{cos}\:{qx}}{{x}^{\mathrm{2}} } \\ $$ Answered by bemath last updated on 06/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:\left(\frac{{p}+{q}}{\mathrm{2}}\:{x}\right)\mathrm{sin}\:\left(\frac{{p}−{q}}{\mathrm{2}}\:{x}\right)}{{x}^{\mathrm{2}} } \\…
Question Number 6703 by love math last updated on 15/Jul/16 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{l}im}\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \left({e}^{{x}} −\mathrm{1}\right)} \\ $$ Answered by FilupSmith last updated on 15/Jul/16 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}}…
Question Number 6689 by Tawakalitu. last updated on 13/Jul/16 $${limit}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}} \\ $$$${x}\:\dashrightarrow\:\infty \\ $$ Answered by FilupSmith last updated on 14/Jul/16 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−\infty} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\infty}…
Question Number 6692 by Tawakalitu. last updated on 14/Jul/16 $${limit}\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}\:−\:\mathrm{1}} \\ $$$${x}\:\rightarrow\:\infty \\ $$ Answered by FilupSmith last updated on 14/Jul/16 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{1}+\frac{\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{1}+\frac{\mathrm{1}}{\infty−\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\infty}…
Question Number 72218 by aliesam last updated on 26/Oct/19 Commented by turbo msup by abdo last updated on 26/Oct/19 $${let}\:{A}\left({x}\right)=\left(\frac{{sinx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow \\ $$$${A}\left({x}\right)={e}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{{sinx}}{{x}}\right)}…
Question Number 137737 by bemath last updated on 06/Apr/21 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{{x}} −\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}} +\mathrm{3}^{−{x}} }\:=? \\ $$$$ \\ $$ Answered by greg_ed last updated on…
Question Number 6668 by janzmir6996 last updated on 10/Jul/16 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}cos}\:\mathrm{2}{x}/\mathrm{sin}\:{x}\mathrm{2} \\ $$ Commented by Rasheed Soomro last updated on 10/Jul/16 $${Do}\:{you}\:{mean}\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\mathrm{cos}\:\mathrm{2}{x}/\mathrm{sin}\:{x}^{\mathrm{2}} \:\:\:? \\…