Question Number 218543 by agmed last updated on 11/Apr/25 Commented by Ghisom last updated on 11/Apr/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} } \\ $$ Answered by SdC355 last…
Question Number 218309 by 200392jjlv last updated on 05/Apr/25 $$\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{4}}{\mathrm{16}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\:\underset{−\infty} {\overset{\mathrm{0}} {\int}}\frac{\mathrm{4}}{\mathrm{16}+{x}^{\mathrm{2}} }{dx}+\underset{\infty} {\int}^{\mathrm{0}} \frac{\mathrm{4}}{\mathrm{16}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\: \\ $$$$…
Question Number 218196 by mnjuly1970 last updated on 01/Apr/25 $$ \\ $$$$ \\ $$$$\:\:\:\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\:\left(\:\frac{\left(\mathrm{2}{n}\right)!}{{n}!}\:\right)^{\frac{\mathrm{1}}{{n}}} =\:?\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Answered by mehdee7396 last updated…
Question Number 217190 by efronzo1 last updated on 05/Mar/25 $$\mathrm{Given}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \:+\:\mathrm{a}_{\mathrm{n}+\mathrm{2}} \: \\ $$$$\:\:\mathrm{where}\:\mathrm{a}_{\mathrm{3}} =\:\mathrm{4}\:\mathrm{and}\:\mathrm{a}_{\mathrm{5}} =\:\mathrm{6} \\ $$$$\:\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \:. \\ $$ Answered by mathmax…
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Question Number 216953 by mustaphapelumi last updated on 25/Feb/25 Commented by Ghisom last updated on 26/Feb/25 $$\mathrm{or}\:\mathrm{use}\:\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}\left[\mathrm{sin}\:{x}\right]}{{dx}}}{\frac{{d}\left[{x}\right]}{{dx}}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}}{\mathrm{1}}\:=\mathrm{1} \\ $$…
Question Number 216926 by Engr_Jidda last updated on 24/Feb/25 $${Evaluate}\:\mathrm{5}^{\mathrm{2}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{m}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{m}^{\mathrm{2}} +\mathrm{2}{m}}\right)^{{n}−\mathrm{1}} \\ $$ Answered by Wuji last updated on 25/Feb/25…
Question Number 216925 by Engr_Jidda last updated on 24/Feb/25 Answered by mehdee7396 last updated on 24/Feb/25 $${s}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{4}{u}+\mathrm{1}\right){du}+\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{4}{u}+\mathrm{1}\right){du}+…+\int_{\mathrm{0}} ^{\mathrm{10}} \left(\mathrm{4}{u}+\mathrm{1}\right){du} \\ $$$$\left.=\left.\left(\left.\mathrm{2}{u}^{\mathrm{2}}…
Question Number 216917 by alcohol last updated on 24/Feb/25 $${li}\underset{{x}\rightarrow+\infty} {{m}}\:\left(\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}−\sqrt{{x}}\right) \\ $$ Answered by mehdee7396 last updated on 24/Feb/25 $${lim}_{{x}\rightarrow\infty} \:\frac{\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}}{\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}+\sqrt{{x}}}} \\ $$$$={lim}_{{x}\rightarrow\infty} \:\frac{\sqrt{{x}}}{\:\mathrm{2}\sqrt{{x}}}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 216748 by sniper237 last updated on 17/Feb/25 $${Find}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} {ln}\mid\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+{it}\right)\mid{dt} \\ $$ Answered by issac last updated on 18/Feb/25 $$\mathrm{can}'\mathrm{t}\:\mathrm{simplify} \\ $$$$\mathrm{function}\:\mathrm{ln}\left(\mid\Gamma\left(\boldsymbol{{i}}{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mid\right) \\…