Question Number 59393 by Mikael_Marshall last updated on 09/May/19 $$\underset{{x}\rightarrow+\infty} {{lim}}\:\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{2}}\overset{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}} {\right)} \\ $$$${pls}. \\ $$ Commented by maxmathsup by imad last…
Question Number 124915 by liberty last updated on 06/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2tan}\:{x}−\mathrm{5sin}\:{x}+\mathrm{3}{x}−{x}^{\mathrm{3}} }{\mathrm{1}−\sqrt[{\mathrm{5}}]{\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} }}\:=?\: \\ $$ Answered by bemath last updated on 07/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)−\mathrm{5}\left({x}−\frac{{x}^{\mathrm{3}}…
Question Number 124802 by john_santu last updated on 06/Dec/20 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}\:−\:\sqrt[{\mathrm{4}}]{{x}}}{\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\:\sqrt[{\mathrm{3}}]{{x}}}\:=?\: \\ $$ Answered by bramlexs22 last updated on 06/Dec/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}}\:\left(\sqrt[{\mathrm{4}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}\right)}{\:\sqrt[{\mathrm{3}}]{{x}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{12}}]{{x}}}\:\left(\frac{\sqrt[{\mathrm{4}\:}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}}\right)\:=\:\mathrm{0}…
Question Number 190317 by krishvasoya last updated on 31/Mar/23 Answered by senestro last updated on 31/Mar/23 $$\frac{\mathrm{1}}{{e}+\mathrm{1}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 190300 by mnjuly1970 last updated on 31/Mar/23 $$ \\ $$$$\:\:\:\mathrm{lim}_{\:{x}\rightarrow\mathrm{0}} \frac{\:{sin}\left({x}\right)\:−{tan}\left({x}\right)}{{x}^{\:\mathrm{3}} } \\ $$$$\:\:\:\:\:{a}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:{c}:\:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:{d}:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$ Answered by som(math1967) last updated…
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Question Number 190182 by SLVR last updated on 29/Mar/23 $${Li}\underset{{x}\rightarrow\pi/\mathrm{2}} {{m}}\frac{{sinx}−{sinx}^{{sinx}} }{\mathrm{1}−{sinx}+{logsinx}} \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\:\:\:\:\:\:\:\:{b}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:{c}\right)\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:{d}\right){none} \\ $$ Commented by SLVR last updated on 29/Mar/23 $${kindly}\:{help}\:{me} \\…
Question Number 190178 by mnjuly1970 last updated on 29/Mar/23 $$ \\ $$$$\:\:{calculate} \\ $$$$ \\ $$$$\:\:\:\:\mathrm{lim}_{\:\mathrm{n}\rightarrow\infty} \frac{\:\Gamma\:\left(\:\frac{\:{n}+\mathrm{3}}{\mathrm{2}}\:\right)}{{n}^{\:\frac{\mathrm{3}}{\mathrm{2}}} .\Gamma\:\left(\frac{{n}}{\mathrm{2}}\:\right)}\:=\:? \\ $$ Commented by Frix last updated…
Question Number 190165 by TUN last updated on 29/Mar/23 $${S}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$=>\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =¿ \\ $$ Answered by JDamian last updated…
Question Number 124555 by mnjuly1970 last updated on 04/Dec/20 Answered by mnjuly1970 last updated on 05/Dec/20 $${solution}: \\ $$$$\:\:{f}\::\left[{a},{b}\right]\rightarrow\mathbb{R}^{+} \cup\left\{\mathrm{0}\right\}\:{is}\:{continuous}\:: \\ $$$$\:\:\:\:{lim}_{{n}\rightarrow\infty} \left(\int_{{a}} ^{\:{b}} {f}^{\:{n}}…