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 65301 by LPM last updated on 28/Jul/19 Commented by MJS last updated on 28/Jul/19 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{ve}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{30}\:\mathrm{because}\:\mathrm{then} \\ $$$$\frac{{ab}}{\mathrm{3}}=−\mathrm{3}\:\mathrm{but}\:\mathrm{with}\:\mathrm{tbe}\:\mathrm{given}\:\mathrm{equations}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{values}\:\mathrm{different}\:\mathrm{from}\:\mathrm{each}\:\mathrm{other} \\ $$…
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 $$\left(\mathrm{1}\right)\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}}…
Question Number 65300 by rajesh4661kumar@gmail.com last updated on 28/Jul/19 Commented by kaivan.ahmadi last updated on 28/Jul/19 $${f}\left({x}+\Delta{x}\right)={f}\left({x}\right)+{f}'\left({x}\right).\Delta{x} \\ $$$${f}\left({x}\right)={logx}\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{1}}{{xln}\mathrm{10}} \\ $$$$\left.{f}\left(\mathrm{10}.\mathrm{02}\right)\right)={f}\left(\mathrm{10}+\mathrm{0}.\mathrm{02}\right)={f}\left(\mathrm{10}\right)+{f}'\left(\mathrm{10}\right)×\mathrm{0}.\mathrm{02}= \\ $$$$\mathrm{2}.\mathrm{3026}+\frac{\mathrm{1}}{\mathrm{10}{ln}\mathrm{10}}×\mathrm{0}.\mathrm{02}=\mathrm{2}.\mathrm{3026}+\frac{\mathrm{2}}{\mathrm{1000}{ln}\mathrm{10}} \\ $$$$…
Question Number 130835 by mathmax by abdo last updated on 29/Jan/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 65297 by mathmax by abdo last updated on 28/Jul/19 $${let}\:\:\:{U}_{{n}} \:\:{a}\:{sequence}\:{wich}\:{verify}\:\:{U}_{{n}} \:+{U}_{{n}+\mathrm{1}} +{U}_{{n}+\mathrm{2}} \:={n}\left(−\mathrm{1}\right)^{{n}} \\ $$$${for}\:{all}\:{integr}\:{n}\:\:\:{calculate}\:{interms}\:{of}\:{n} \\ $$$${A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{U}_{{k}} \\…
Question Number 130830 by Ar Brandon last updated on 29/Jan/21 $$\mathrm{x}^{\mathrm{2}} \mathrm{y}''−\mathrm{xy}'+\mathrm{y}=\mathrm{0} \\ $$ Answered by bemath last updated on 29/Jan/21 $$\mathrm{Cauchy}−\mathrm{Euler}\:\mathrm{diff}\:\mathrm{eq} \\ $$$$\mathrm{let}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{m}} \:\rightarrow\begin{cases}{\mathrm{y}'=\mathrm{mx}^{\mathrm{m}−\mathrm{1}}…
Question Number 65293 by mathmax by abdo last updated on 27/Jul/19 $$\left.\mathrm{1}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}−{a}}\:\:{with}\:{a}\:\in{C} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+\mathrm{1}}\:\:{and}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{6}} \:+\mathrm{1}} \\ $$$${by}\:{using}\:{the}\:{decomposition}\:{inside}\:{C}\left({x}\right). \\…
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Question Number 130827 by bemath last updated on 29/Jan/21 $$\left[\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}\:} \mathrm{D}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\mathrm{D}+\mathrm{1}\:\right]\mathrm{y}\:=\:\mathrm{4cos}\:\left(\mathrm{ln}\left(\:\mathrm{x}+\mathrm{1}\right)\right) \\ $$ Answered by EDWIN88 last updated on 29/Jan/21 $${let}\:\mathrm{ln}\:\left({x}+\mathrm{1}\right)={t}\:\Rightarrow{x}+\mathrm{1}\:=\:{e}^{{t}} \\ $$$$\begin{cases}{\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}.\frac{{dy}}{{dt}}}\\{\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:=\:\frac{{d}}{{dx}}\:\left[\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\frac{{dy}}{{dt}}\:\right]=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}}…