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Question Number 84130 by mahdi last updated on 09/Mar/20 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}x}\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}}−\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}+\mathrm{x}\right) \\ $$ Commented by MJS last updated on 10/Mar/20 $$\mathrm{let}\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}^{+}…
Question Number 18593 by Joel577 last updated on 25/Jul/17 Answered by 433 last updated on 25/Jul/17 $$ \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{2016}\left(\frac{\mathrm{1}^{\mathrm{2015}} }{{n}^{\mathrm{2014}} }+\frac{\mathrm{2}^{\mathrm{2015}}…
Question Number 149660 by fotosy2k last updated on 06/Aug/21 Answered by JDamian last updated on 06/Aug/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{a}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }+\frac{{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }\right)= \\…
Question Number 18587 by Joel577 last updated on 25/Jul/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{{x}\:\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\:+\:\mathrm{1}} \\ $$ Commented by Joel577 last updated on 25/Jul/17 $$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{{x}\:\mathrm{sin}\:{x}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{\mathrm{sin}\:{x}\:\left({x}\:+\:\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\…
Question Number 149637 by mnjuly1970 last updated on 06/Aug/21 Answered by iloveisrael last updated on 06/Aug/21 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{x}^{−\mathrm{1}} +\mathrm{x}^{−\mathrm{4}} }\:−\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{−\mathrm{1}} \right)}{\mathrm{x}} \\ $$$$=\:\underset{{x}\rightarrow\infty}…
Question Number 84083 by john santu last updated on 09/Mar/20 $$\underset{\mathrm{a}\rightarrow\mathrm{x}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{x}−\mathrm{a}\:}\right)}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:= \\ $$ Commented by mathmax by abdo last updated on 09/Mar/20…
Question Number 149596 by iloveisrael last updated on 06/Aug/21 $$\:\mathrm{Without}\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\:\underset{{x}\rightarrow\pi/\mathrm{7}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{sin}\:\mathrm{3x}−\frac{\sqrt{\mathrm{7}}}{\mathrm{8}}}{\mathrm{x}−\frac{\pi}{\mathrm{7}}}\:=? \\ $$ Answered by EDWIN88 last updated on 07/Aug/21 $$\: \\ $$$$\mathrm{let}\:{x}−\frac{\pi}{\mathrm{7}}={t}\:;\:{x}={t}+\frac{\pi}{\mathrm{7}}\:\wedge\:{t}\rightarrow\mathrm{0}…
Question Number 149547 by iloveisrael last updated on 06/Aug/21 $$\:\mathrm{Evaluate}\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{7}}} {\mathrm{lim}}\frac{\mathrm{8cos}\:\mathrm{xcos}\:\mathrm{2xcos}\:\mathrm{4x}+\mathrm{1}}{\mathrm{x}−\frac{\pi}{\mathrm{7}}} \\ $$ Answered by EDWIN88 last updated on 06/Aug/21 Terms of Service Privacy Policy…
Question Number 83943 by jagoll last updated on 08/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:\mathrm{6x}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{2x}\:\mathrm{tan}\:\mathrm{5x}} \\ $$$$ \\ $$ Answered by jagoll last updated on 08/Mar/20 Terms of Service…