Question Number 160394 by qaz last updated on 29/Nov/21 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }} \left(\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{3}} −\mathrm{1}}\right)^{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{n}−\mathrm{2}}} } \centerdot…\centerdot\left(\frac{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }{\mathrm{2}^{\mathrm{n}} −\mathrm{1}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =? \\ $$ Terms of…
Question Number 160384 by qaz last updated on 28/Nov/21 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{n}} }\mathrm{dx}=? \\ $$ Answered by mathmax by abdo last updated on 29/Nov/21…
Question Number 29310 by tawa tawa last updated on 07/Feb/18 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}:\:\:\:\:\:\left\{\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\right\}_{\mathrm{n}\:\:=\:\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\:\mathrm{is}\:\mathrm{convergent}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160330 by alcohol last updated on 27/Nov/21 $$\frac{{du}}{{dx}}\:=\:{e}^{\left(\frac{{x}}{{u}}\right)} \\ $$$${find}\:{u} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94764 by Hamida last updated on 20/May/20 Answered by prakash jain last updated on 20/May/20 $$\mathrm{1a}\:\left(\mathrm{i}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}=\mathrm{1}\:\left(\mathrm{top}\:\mathrm{point}\:\mathrm{of}\:\mathrm{semicircle}\right) \\ $$$$\mathrm{1a}\left(\mathrm{ii}\right) \\ $$$$\underset{{x}\rightarrow−\mathrm{1}^{−}…
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Question Number 94730 by i jagooll last updated on 20/May/20 Answered by john santu last updated on 20/May/20 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\mathrm{e}^{\mathrm{ln}\left(\mathrm{x}\right)^{\mathrm{sin}\:\mathrm{x}} } =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{e}^{\mathrm{sin}\:\mathrm{x}.\:\mathrm{ln}\left(\mathrm{x}\right)} \\…
Question Number 160256 by qaz last updated on 26/Nov/21 $$\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}^{\mathrm{x}} −\mathrm{sin}\:\mathrm{2}^{\mathrm{x}} }{\mathrm{2}^{\mathrm{x}^{\mathrm{x}} } −\mathrm{2}^{\mathrm{2}^{\mathrm{x}} } }=? \\ $$ Commented by cortano last updated on…
Question Number 94723 by john santu last updated on 20/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\int}}\:\sqrt{\mathrm{4}+{t}^{\mathrm{3}} \:}\:{dt}}{{x}^{\mathrm{2}} }\:?\: \\ $$ Answered by i jagooll last updated…
Question Number 29180 by A1B1C1D1 last updated on 05/Feb/18 $$\underset{\mathrm{x}\:\rightarrow\:\mathrm{a}} {\mathrm{lim}}\:\left(\frac{\sqrt[{\mathrm{m}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{m}}]{\mathrm{a}}}{\:\sqrt[{\mathrm{n}}]{\mathrm{x}}\:−\:\sqrt[{\mathrm{n}}]{\mathrm{a}}}\right) \\ $$$$\left.\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{L}'\mathrm{hospital}\:\mathrm{rules}\::-\right) \\ $$ Answered by Rasheed.Sindhi last updated on 05/Feb/18 $$\mathrm{Formula} \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{x}\rightarrow\mathrm{a}}…