Question Number 118327 by bramlexs22 last updated on 17/Oct/20 $$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}−\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{\mathrm{2}{x}−\sqrt{\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} }}\:? \\ $$ Answered by TANMAY PANACEA last updated on 17/Oct/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}}…
Question Number 118272 by bramlexs22 last updated on 16/Oct/20 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{{a}}{\mathrm{1}−{x}^{{a}} }\:−\:\frac{{b}}{\mathrm{1}−{x}^{{b}} }\:\right)\:=\: \\ $$$${with}\:\left({a},{b}\right)\:\in\left(\mathbb{R}^{+} \right)^{\mathrm{2}} \: \\ $$ Answered by mathmax by abdo last…
Question Number 118247 by bemath last updated on 16/Oct/20 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{4}} }\:−\:\sqrt[{\mathrm{5}}]{\mathrm{1}+{x}^{\mathrm{5}} }\:=? \\ $$ Answered by bobhans last updated on 16/Oct/20 $${solve}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{4}} }\:−\:\sqrt[{\mathrm{5}\:}]{\mathrm{1}+{x}^{\mathrm{5}}…
Question Number 118228 by bobhans last updated on 16/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sum_{\mathrm{r}=\mathrm{1}} ^{\mathrm{10}} \left(\mathrm{x}+\mathrm{r}\right)^{\mathrm{2020}} }{\left(\mathrm{x}^{\mathrm{1008}} +\mathrm{1}\right)\left(\mathrm{3x}^{\mathrm{1012}} +\mathrm{1}\right)}\:=?\: \\ $$ Answered by MJS_new last updated on 16/Oct/20…
Question Number 118140 by bemath last updated on 15/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left[\:{x}^{\mathrm{2}} \:\right]}{\mathrm{2}{x}}\:=?\: \\ $$ Answered by TANMAY PANACEA last updated on 15/Oct/20 $${let}\:{x}=\mathrm{0}.\mathrm{9}\:\:{x}^{\mathrm{2}} =\mathrm{0}.\mathrm{81}\:\:\left[\mathrm{0}.\mathrm{81}\right]=\mathrm{0} \\…
Question Number 183673 by universe last updated on 28/Dec/22 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\:\left[\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{{a}\mathrm{x}}}\right)\right]^{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2}−\boldsymbol{\mathrm{bx}}}\right)} \\ $$ Answered by AlexandreT last updated on 28/Dec/22 $${lim}\:\left[{sen}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{ax}}\right)−\mathrm{1}\right]{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}−{bx}}\right)…
Question Number 118088 by bobhans last updated on 15/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}.\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:.\sqrt[{\mathrm{3}\:}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{3}} }\:=? \\ $$ Answered by bemath last updated on 15/Oct/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{x}.\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 118040 by mathocean1 last updated on 14/Oct/20 Answered by Ar Brandon last updated on 14/Oct/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2cos}{x}}{\pi−\mathrm{3}{x}}=\frac{\mathrm{0}}{\mathrm{0}}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\frac{\mathrm{2sin}{x}}{−\mathrm{3}}=\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\boldsymbol{\div}−\mathrm{3}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$ Answered by bobhans…
Question Number 183528 by universe last updated on 26/Dec/22 $$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\:\mathrm{3}\sqrt{\mathrm{1}+\frac{\mathrm{k}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} }}\:−\mathrm{1}\right)\:\:=\:\:\:? \\ $$ Commented by TANMAY PANACEA last updated…
Question Number 183483 by cortano1 last updated on 26/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\sqrt{{x}}}{\:\sqrt{{x}−\sqrt{{x}−\sqrt{{x}−\sqrt{{x}−…}}}}}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com