Question Number 153517 by liberty last updated on 08/Sep/21 $$\:\:{L}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt{\mathrm{2019}{x}−\mathrm{2018}}−\mathrm{1}}{\:\sqrt[{\mathrm{2018}}]{{x}^{\mathrm{2019}} }−\mathrm{1}} \\ $$$$\:{then}\:\mathrm{2}×{L}\:=? \\ $$ Commented by MJS_new last updated on 08/Sep/21 $$\mathrm{just}\:“\mathrm{hospitalize}''\:\mathrm{it}\:\Rightarrow\:{L}=\mathrm{1009} \\…
Question Number 153518 by Tawa11 last updated on 08/Sep/21 $$\mathrm{Show}\:\mathrm{whether}\:\:\:\underset{\mathrm{n}\:\:=\:\:\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}\:\:\:\:+\:\:\:\mathrm{n}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }\right)\:\:\:\:\:\mathrm{is}\:\mathrm{uniformly}\:\mathrm{convegence}\:\mathrm{for}\:\mathrm{real} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$ Answered by puissant last updated on…
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Question Number 153449 by liberty last updated on 07/Sep/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)}−\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{3}{x}\right)}}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}−\sqrt{\mathrm{1}+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22365 by A1B1C1D1 last updated on 16/Oct/17 Answered by sma3l2996 last updated on 16/Oct/17 $$=\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} }}{\mathrm{1}+\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}^{{x}} {x}}}=\frac{\mathrm{1}−\mathrm{0}+\mathrm{0}}{\mathrm{1}+\mathrm{0}+\mathrm{0}}=\mathrm{1}\:\: \\ $$ Commented…
Question Number 153420 by alcohol last updated on 07/Sep/21 $${how}\:{many}\:{x}\:\in\mathbb{R}\:{satisfy}\:{x}^{\mathrm{99}} −\mathrm{99}{x}+\mathrm{1}=\mathrm{0} \\ $$ Answered by mr W last updated on 07/Sep/21 $${f}'\left({x}\right)=\mathrm{99}{x}^{\mathrm{98}} −\mathrm{99}=\mathrm{0} \\ $$$${x}=\pm\mathrm{1}…
Question Number 153404 by liberty last updated on 07/Sep/21 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{8}^{{x}} +\mathrm{3}^{{x}} }\:−\sqrt{\mathrm{4}^{{x}} −\mathrm{2}^{{x}} }\:=? \\ $$ Answered by EDWIN88 last updated on 07/Sep/21 $$\:\underset{{x}\rightarrow\infty}…
Question Number 153352 by liberty last updated on 06/Sep/21 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{27}^{{x}} +\mathrm{9}^{{x}} }\:−\sqrt{\mathrm{9}^{{x}} +\mathrm{3}^{{x}} }\:=? \\ $$ Answered by MJS_new last updated on 06/Sep/21 $$\underset{{x}\rightarrow+\infty}…
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Question Number 153346 by mathlove last updated on 06/Sep/21 Answered by liberty last updated on 06/Sep/21 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−\mathrm{4}}{{x}−\mathrm{2}}=\mathrm{15}\:\rightarrow\begin{cases}{{f}\left(\mathrm{2}\right)=\mathrm{4}}\\{{f}\:'\left(\mathrm{2}\right)=\mathrm{15}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}{x}^{\mathrm{2}} −{f}\left({x}\right)}{{x}−\mathrm{2}}\:=\infty \\ $$ Answered…