Question Number 99340 by bobhans last updated on 20/Jun/20 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{2x}−\mathrm{a}\:,\:\mathrm{x}<−\mathrm{3}}\\{\mathrm{ax}+\mathrm{b}\:,\:−\mathrm{3}\leqslant\mathrm{x}\leqslant\mathrm{3}}\\{\mathrm{b}−\mathrm{5x}\:,\:\mathrm{x}>\mathrm{3}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}f}\left(\mathrm{x}\right)\:\mathrm{and}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{exist}\: \\ $$ Commented by PRITHWISH SEN 2 last updated…
Question Number 99262 by Ar Brandon last updated on 19/Jun/20 $$\mathrm{Consider}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{xe}^{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{Find}\:\mathrm{3}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\:\mathrm{b},\:\mathrm{and}\:\mathrm{c}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{ax}+\mathrm{b}+\frac{\mathrm{c}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}}\xi\left(\mathrm{x}\right) \\ $$$$\mathrm{where}\:\underset{\mathrm{x}\rightarrow\pm\infty} {\mathrm{lim}}\xi\left(\mathrm{x}\right)=\mathrm{0} \\ $$ Terms of Service Privacy…
Question Number 99244 by Ar Brandon last updated on 19/Jun/20 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}−\mathrm{4}}−\mathrm{2}}{\mathrm{2x}}=? \\ $$ Answered by mahdi last updated on 19/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}+\mathrm{4}}−\mathrm{2}}{\mathrm{2x}}×\frac{\sqrt{\mathrm{x}+\mathrm{4}}+\mathrm{2}}{\:\sqrt{\mathrm{x}−\mathrm{4}}+\mathrm{2}}= \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 99193 by bemath last updated on 19/Jun/20 Commented by I want to learn more last updated on 19/Jun/20 $$\mathrm{Sir},\:\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{type}\:\mathrm{like}\:\mathrm{this} \\ $$ Commented by…
Question Number 164705 by Kayela last updated on 20/Jan/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}^{{x}} +\mathrm{2}^{{x}} +…+{n}^{{x}} }{{n}}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered by Berlindo last updated on 20/Jan/22 $$=\left[\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}^{\mathrm{x}}…
Question Number 164702 by mathls last updated on 20/Jan/22 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left(\frac{\mathrm{1}^{{x}} +\mathrm{2}^{{x}} +\centerdot\centerdot\centerdot+{n}^{{x}} }{{n}}\right)^{\frac{\mathrm{1}}{{x}}} =? \\ $$ Answered by mahdipoor last updated on 21/Jan/22 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 33595 by abdo imad last updated on 19/Apr/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left(\mathrm{1}+{sinx}\right)\:−{x}\sqrt{\mathrm{1}−{x}}}{{sinx}\:−{shx}}\:\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33593 by abdo imad last updated on 19/Apr/18 $${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{2}\left(\mathrm{1}−{cosx}\right){sinx}\:−{x}^{\mathrm{3}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{sin}^{\mathrm{5}} {x}\:−{x}^{\mathrm{5}} } \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 33594 by abdo imad last updated on 19/Apr/18 $${calculate}\:\:{lim}_{{x}\rightarrow\mathrm{1}^{−} } \:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\alpha} }\left({arcsinx}\:−\frac{\pi}{\mathrm{2}}\right)\:. \\ $$ Commented by abdo imad last updated on 24/Apr/18 $${let}\:{use}\:{the}\:{ch}.\:\mathrm{1}−{x}\:={t}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}^{−}…
Question Number 164585 by mnjuly1970 last updated on 19/Jan/22 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathcal{I}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{Li}_{\:\mathrm{2}} \:\left(\:{x}\:\right)}{\mathrm{1}\:+\:{x}}\:{dx}\:=\:? \\ $$$$\:\:\:\:−−−−−−\: \\ $$ Answered by mindispower last updated on…