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tinkutara-admint-I-want-to-update-to-version-2-074-

Question Number 95420 by bobhans last updated on 25/May/20 $$\mathrm{tinkutara}\:\mathrm{admint} \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{update}\:\mathrm{to}\:\mathrm{version}\:\mathrm{2}.\mathrm{074} \\ $$ Commented by Tinku Tara last updated on 25/May/20 Version uploaded so far on playstore is 2.073. 2.076 is only available on www.tinkutara.com and contains fixes for problem reported by MrW. Commented by…

Calculate-a-lim-x-x-1-x-1-x-2-b-lim-x-0-2-1-x-1-2-1-x-1-

Question Number 160917 by LEKOUMA last updated on 09/Dec/21 $${Calculate} \\ $$$$\left.{a}\right)\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{{x}^{\mathrm{2}} } \\ $$$$\left.{b}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{1}}{{x}}} +\mathrm{1}} \\ $$ Answered by jama last…

Calculate-1-lim-x-0-x-1-2x-1-x-2-lim-x-a-sin-x-sin-a-1-x-a-

Question Number 160900 by LEKOUMA last updated on 08/Dec/21 $${Calculate} \\ $$$$\left.\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}+\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}\right)^{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{a}}\right)^{\frac{\mathrm{1}}{{x}−{a}}} \\ $$ Commented by cortano last updated on…