Question Number 144564 by imjagoll last updated on 26/Jun/21 Commented by justtry last updated on 26/Jun/21 $$\mathrm{0}?? \\ $$ Answered by liberty last updated on…
Question Number 144556 by imjagoll last updated on 26/Jun/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\right)\:=? \\ $$ Answered by liberty last updated on 26/Jun/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\sqrt{\mathrm{cos}\:\mathrm{2x}}}{\mathrm{x}^{\mathrm{2}} }\right)=? \\…
Question Number 144558 by imjagoll last updated on 26/Jun/21 $$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}−\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}}}{\mathrm{x}−\mathrm{1}}\right)=? \\ $$ Answered by liberty last updated on 26/Jun/21 $$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}−\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}}}{\mathrm{x}−\mathrm{1}}\right)=? \\…
Question Number 79010 by jagoll last updated on 22/Jan/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{3x}\:\mathrm{ln}\left(\mathrm{x}\right)\:? \\ $$ Answered by jagoll last updated on 22/Jan/20 Commented by jagoll last updated…
Question Number 144513 by mnjuly1970 last updated on 26/Jun/21 Answered by Olaf_Thorendsen last updated on 26/Jun/21 $$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{−{x}} \:=\:{e}^{−{x}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)} \:\underset{\infty} {\sim}\:{e}^{−{x}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right)} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{−{x}} \:\underset{\infty} {\sim}\:\frac{\mathrm{1}}{{e}}{e}^{\frac{\mathrm{1}}{\mathrm{2}{x}}}…
Question Number 144502 by alcohol last updated on 26/Jun/21 $${it}\:{is}\:{known}\:{that}\:{after}\:{injecting}\:{the}\:\mathrm{1}^{{st}} \\ $$$${dosage}\:{of}\:{a}\:{drug}\:{in}\:{exactly}\:{hours} \\ $$$${p}\left({t}\right)=\begin{cases}{{Ae}^{−\frac{{t}}{\mathrm{24}}} ,\:\mathrm{0}\leqslant{t}\leqslant\mathrm{8}}\\{\left({A}+{Ae}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right){e}^{−\left(\frac{{t}−\mathrm{8}}{\mathrm{24}}\right)} ,\:\mathrm{8}\:\leqslant\:{t}\leqslant\mathrm{24}}\end{cases} \\ $$$${A}\:{is}\:{the}\:{initial}\:{dosage}\:{at}\:{t}=\:\mathrm{0}{hours} \\ $$$$\left.{a}\right)\:{show}\:{that}\:{towards}\:{the}\:{application}\:{of} \\ $$$${the}\:{n}^{{th}} \:{dosage}\:,\:{the}\:{dosage}\:{remained}\:{in}\:{blood}\:{is} \\…
Question Number 144471 by pyra last updated on 25/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 144447 by pyra last updated on 25/Jun/21 Commented by Canebulok last updated on 25/Jun/21 indeterminate Answered by Dwaipayan Shikari last updated on 25/Jun/21…
Question Number 144413 by imjagoll last updated on 25/Jun/21 $$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4}\sqrt{\mathrm{x}}}−\frac{\mathrm{1}}{\mathrm{5}−\mathrm{5}\sqrt[{\mathrm{5}}]{\mathrm{x}}}\:\right]\:=? \\ $$ Answered by Olaf_Thorendsen last updated on 25/Jun/21 $${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4}\sqrt{{x}}}−\frac{\mathrm{1}}{\mathrm{5}−\mathrm{5}\sqrt[{\mathrm{5}}]{{x}}} \\ $$$$\mathrm{Let}\:{x}\:=\:{X}^{\mathrm{10}} \\ $$$${f}\left({X}^{\mathrm{10}}…
Question Number 144406 by pyra last updated on 25/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com