Question Number 145129 by liberty last updated on 02/Jul/21 Answered by EDWIN88 last updated on 03/Jul/21 $$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{1}\right)}{\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{1}\right)\left(\mathrm{2sin}^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{2}{x}}}{\mathrm{2}}\right)\right)^{{n}} \:}\:=\:{a}…
Question Number 79565 by jagoll last updated on 26/Jan/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}\:\right)×\left(\frac{\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}} +\mathrm{x}\right)}{\mathrm{x}}\right) \\ $$ Commented by john santu last updated on 26/Jan/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 145065 by Ar Brandon last updated on 02/Jul/21 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}}}=\mathrm{1}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{2}^{\mathrm{2k}} }\mathrm{C}_{\mathrm{2k}} ^{\mathrm{k}} \right]\mathrm{x}^{\mathrm{k}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{n}} \right) \\ $$ Terms of Service Privacy…
Question Number 79499 by jagoll last updated on 25/Jan/20 $$\underset{\mathrm{0}} {\overset{\mathrm{30}\pi} {\int}}\mid\mathrm{sin}\:\mathrm{x}\mid\:\mathrm{dx}=\: \\ $$ Commented by john santu last updated on 25/Jan/20 $${y}\:=\:\mid\mathrm{sin}\:{x}\mid\:{is}\:{even}\:{function}\:{and} \\ $$$${periodic}\:{with}\:{periode}\:=\:\pi…
Question Number 145012 by mim24 last updated on 01/Jul/21 Answered by ArielVyny last updated on 01/Jul/21 $${lim}\left({x}^{−\frac{\mathrm{1}}{{x}}} \right)\:\:\left({x}\rightarrow\mathrm{0}\right) \\ $$$${lim}\left({e}^{−\frac{\mathrm{1}}{{x}}{lnx}} \right)\left({x}\rightarrow\mathrm{0}^{+} \right)=+\infty \\ $$$$ \\…
Question Number 145009 by mim24 last updated on 01/Jul/21 Answered by ArielVyny last updated on 01/Jul/21 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{xcosx}}{{sinx}}+\frac{{xcos}\left(\mathrm{2}{x}\right)}{{sinx}} \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{x}}{{tgx}}+\frac{{x}}{{sinx}}\left(\mathrm{2}{sin}^{\mathrm{2}} {x}+\mathrm{1}\right)=\frac{{x}}{{tgx}}+\mathrm{2}{xsinx}+\frac{{x}}{{sinx}} \\ $$$${lim}\left(\mathrm{2}{xsinx}\right)=\mathrm{0}\left({x}\rightarrow\mathrm{0}\right) \\…
Question Number 145011 by mim24 last updated on 01/Jul/21 Answered by ArielVyny last updated on 01/Jul/21 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{{ln}\left(\mathrm{1}+{x}\right)}={lim}\frac{\mathrm{2}{e}^{−\mathrm{2}{x}} }{\frac{\mathrm{1}}{\mathrm{1}+{x}}}={lim}\left(\mathrm{2}{e}^{−\mathrm{2}{x}} \right)=\mathrm{2} \\ $$$${we}\:{havd}\:{used}\:{hospital}\:{theorem}'{s} \\ $$$$…
Question Number 145010 by mim24 last updated on 01/Jul/21 Answered by ArielVyny last updated on 01/Jul/21 $${consider}\:{f}\left({x}\right)={a}^{{x}} \:\:\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${then}\:{f}'\left({x}\right)={ln}\left({a}\right){a}^{{x}} \\ $$$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{a}^{{x}} −\mathrm{1}}{{x}}={lim}\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}={f}'\left(\mathrm{0}\right)={lna} \\…
Question Number 79443 by jagoll last updated on 25/Jan/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\left(\mathrm{sin}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} +\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{sin}\:\mathrm{x}} \right]\:= \\ $$ Commented by mathmax by abdo last updated on 25/Jan/20 $${let}\:{f}\left({x}\right)=\left({sinx}\right)^{\frac{\mathrm{1}}{{x}}}…
Question Number 144946 by mnjuly1970 last updated on 30/Jun/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\::=\:\int_{\:\mathrm{0}} ^{\:\:\frac{\pi}{\mathrm{2}}} \left(\frac{\:\mathrm{x}}{\mathrm{cot}\:\left(\:\mathrm{x}\:\right)}\:\right)^{\:\mathrm{3}} \mathrm{dx}=? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…