Question Number 141858 by gsk2684 last updated on 24/May/21 Commented by gsk2684 last updated on 24/May/21 $${solution}\:{please} \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 141779 by iloveisrael last updated on 23/May/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{x}}}\:=? \\ $$ Answered by Willson last updated on 23/May/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2x}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}}}}…
Question Number 76191 by abdomathmax last updated on 25/Dec/19 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{e}^{{x}} −{e}^{\left[{x}\right]} }{{x}} \\ $$ Answered by Rio Michael last updated on 25/Dec/19 $${Am}\:{not}\:{sure}\:{if}\:{this}\:{limit} \\…
Question Number 141699 by cesarL last updated on 22/May/21 $${If}\:{lim}_{{x}\rightarrow\mathrm{1}} =\frac{\sqrt[{{k}}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}={L}\neq\mathrm{0}\:\:{Find}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt[{{k}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$ Answered by iloveisrael last updated on 22/May/21 $${Given}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{{k}\:}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}\:=\:{L}\:{equivalent} \\ $$$${to}\:\underset{{t}\rightarrow\mathrm{1}}…
Question Number 76151 by Rio Michael last updated on 24/Dec/19 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{{k}} {e}^{−\mathrm{4}{x}} ,\:{k}>\mathrm{0} \\ $$ Commented by kaivan.ahmadi last updated on 24/Dec/19 $${lim}_{{x}−\rightarrow\infty} \frac{{x}^{{k}}…
Question Number 141627 by cherokeesay last updated on 21/May/21 Answered by MJS_new last updated on 21/May/21 $$\frac{\mathrm{sin}\:\mathrm{2}{x}\:+\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{4}{x}}=\frac{−\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{4cos}^{\mathrm{2}} \:{x}\right)}{\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \:{x}\right)}= \\ $$$$\:\:\:\:\:\left[\mathrm{sin}\:{x}\:=\frac{\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}\wedge\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}\right] \\…
Question Number 141509 by I want to learn more last updated on 19/May/21 $$\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{sin}^{\mathrm{3}} \mathrm{3}\theta\:\:\:−\:\:\:\mathrm{3}\theta^{\mathrm{3}} }{\theta^{\mathrm{3}} \:\:+\:\:\theta^{\mathrm{4}} } \\ $$ Answered by bramlexs22 last…
Question Number 75939 by ahmadshahhimat775@gmail.com last updated on 21/Dec/19 Commented by JDamian last updated on 21/Dec/19 $$\left.{c}\right)\:\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} \\ $$ Commented by mathmax by abdo last…
Question Number 75931 by Rio Michael last updated on 21/Dec/19 $${Evaluate} \\ $$$${a}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{27}{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${b}.\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{x}−\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${c}.\underset{{x}\rightarrow\infty} {\:\mathrm{lim}}\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}−\mathrm{3}} \\…
Question Number 75930 by Rio Michael last updated on 21/Dec/19 $${consider}\:{the}\:{function} \\ $$$$\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} ,\:{if}\:{x}\:{is}\:{rational}}\\{\mathrm{1}\:+\:{x}^{\mathrm{4}} ,\:{if}\:{x}\:{is}\:{irrational}}\end{cases} \\ $$$${Use}\:{the}\:{sandwich}\left({pinchin}\right)\:{theorem}\:{to} \\ $$$${prove}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\mathrm{1}. \\ $$ Terms of Service…