Question Number 9488 by Joel575 last updated on 10/Dec/16 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}\:+\:\mathrm{4}\sqrt{\mathrm{2016}}\:+\:\mathrm{9}\sqrt{\mathrm{2016}}\:+\:…\:+\:{x}^{\mathrm{2}} \sqrt{\mathrm{2016}}}{{x}^{\mathrm{3}} } \\ $$ Answered by mrW last updated on 10/Dec/16 $$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2016}}×\left(\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{2}}…
Question Number 75013 by necxxx last updated on 05/Dec/19 $${The}\:{largest}\:{interval}\:{for}\:{which} \\ $$$${x}^{\mathrm{12}} −{x}^{\mathrm{9}} +{x}^{\mathrm{4}} −{x}+\mathrm{1}>\mathrm{0}\:{is} \\ $$$$\left({a}\right)−\mathrm{4}<{x}\leqslant\mathrm{0} \\ $$$$\left({b}\right)\mathrm{0}<{x}<\mathrm{1} \\ $$$$\left({c}\right)−\mathrm{100}<{x}<\mathrm{100} \\ $$$$\left({d}\right)−\infty<{x}<\infty \\ $$…
Question Number 140530 by liberty last updated on 09/May/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}+\mathrm{y}\right)\mathrm{sec}\:\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{ysec}\:\mathrm{y}}{\mathrm{x}}=? \\ $$ Answered by EDWIN88 last updated on 09/May/21 $$\:\mathrm{L}'\mathrm{H}\ddot {\mathrm{o}pital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:\left(\mathrm{x}+\mathrm{y}\right)+\left(\mathrm{x}+\mathrm{y}\right)\mathrm{sec}\:\left(\mathrm{x}+\mathrm{y}\right)\mathrm{tan}\:\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{0}}{\mathrm{1}}…
Question Number 9439 by tawakalitu last updated on 08/Dec/16 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:,\:\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$$$\left(\mathrm{x}\:+\:\mathrm{1}\right)\left(\mathrm{y}\:+\:\mathrm{3}\right)\:=\:\mathrm{8}\:\:\:…….\:\left(\mathrm{i}\right) \\ $$$$\left(\mathrm{y}\:+\:\mathrm{3}\right)\left(\mathrm{z}\:−\:\mathrm{1}\right)\:=\:\mathrm{3}\:\:………\:\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{z}\:−\:\mathrm{1}\right)\left(\mathrm{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{2}\:\:\:………\:\left(\mathrm{iii}\right) \\ $$ Commented by RasheedSoomro last updated on 09/Dec/16…
Question Number 140471 by EDWIN88 last updated on 08/May/21 $$\mathrm{If}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{a}\:\mathrm{sin}\:\mathrm{bx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:=\:\mathrm{e}^{\mathrm{2}} \\ $$$$\:\begin{cases}{\mathrm{a}=?}\\{\mathrm{b}=?}\end{cases} \\ $$ Answered by benjo_mathlover last updated on 08/May/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:\mathrm{x}+\:\mathrm{a}\:\mathrm{sin}\:\mathrm{bx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}}…
Question Number 74870 by vishalbhardwaj last updated on 02/Dec/19 $$\mathrm{solve}\:\mathrm{with}\:\mathrm{explanation} \\ $$$$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}^{−} } {\mathrm{m}}\left[\frac{\mathrm{x}}{\mathrm{sinx}}\right],\:\mathrm{where}\:\left[\:\:\right]\:\mathrm{represents}\:\mathrm{greatest}\:\mathrm{integer} \\ $$ Commented by mathmax by abdo last updated on 02/Dec/19…
Question Number 140381 by Willson last updated on 07/May/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{xe}^{\mathrm{1}−\mathrm{x}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\:\:\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\sqrt{\mathrm{n}}\:\underset{\mathrm{0}} {\int}^{\:\mathrm{1}} \left[\mathrm{f}\left(\mathrm{x}\right)\right]^{\mathrm{n}} \:\mathrm{dt}\:=\:\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74776 by chess1 last updated on 30/Nov/19 Answered by mind is power last updated on 01/Dec/19 $$\mathrm{we}\:\mathrm{show} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{n}} \left(\frac{\mathrm{tan}^{−} \left(\mathrm{x}\right)}{\mathrm{tan}^{−} \left(\mathrm{x}\right)−\mathrm{x}}\right)^{\mathrm{2}}…
Question Number 74766 by chess1 last updated on 30/Nov/19 Commented by mathmax by abdo last updated on 30/Nov/19 $${changement}\:\frac{\mathrm{1}}{{x}}={t}\:{lead}\:{yo}\:{lim}_{{t}\rightarrow+\infty} \sqrt{{t}+\sqrt{{t}+\sqrt{{t}}}}\:−\sqrt{{t}−\sqrt{{t}+\sqrt{{t}}}} \\ $$$$={lim}_{{t}\rightarrow+\infty} {g}\left({t}\right)\:\:{we}\:{have}\:\:\sqrt{{t}+\sqrt{{t}}}=\sqrt{{t}}×\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{t}}}}\sim\sqrt{{t}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}}}\right) \\ $$$$=\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\sqrt{{t}+\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}+\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}}\:{also}\:\sqrt{{t}−\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}−\sqrt{{t}}−\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 9157 by tawakalitu last updated on 21/Nov/16 $$\underset{{x}\rightarrow\mathrm{y}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{n}} \:−\:\mathrm{y}^{\mathrm{n}} }{\mathrm{x}\:−\:\mathrm{y}} \\ $$ Answered by prakash jain last updated on 21/Nov/16 $${x}^{{n}} −{y}^{{n}}…