Question Number 181173 by universe last updated on 22/Nov/22 Commented by CElcedricjunior last updated on 22/Nov/22 $$\left.\boldsymbol{{a}}\right)\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)= \\ $$$$\left.\boldsymbol{{b}}\right)\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)= \\ $$$$\left.\boldsymbol{{c}}\right)\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{log}}\left(\boldsymbol{{e}}\right) \\ $$ Terms of…
Question Number 115621 by sachin1221 last updated on 27/Sep/20 $$ \\ $$$$\:\:\:\:\:\:\:…\:{advanced}\:\:\:{calculus}…\: \\ $$$$\:\:\:\:\:\:\:{evaluate}\::: \\ $$$${show}\:{that}\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\left[{cos}^{\mathrm{2}{p}} \frac{\pi}{\mathrm{2}{n}}+{cos}^{\mathrm{2}{p}} \frac{\mathrm{2}\pi}{\mathrm{2}{n}}+{cos}^{\mathrm{2}{p}} \frac{\mathrm{3}\pi}{\mathrm{2}{n}}……{cos}^{\mathrm{2}{p}} \frac{\pi}{\mathrm{2}}\right]\:=\underset{{r}=\mathrm{1}} {\overset{{p}} {\prod}}\frac{{p}+{r}}{\mathrm{4}{r}} \\ $$…
Question Number 115616 by bobhans last updated on 27/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{8}\:}]{\mathrm{256}+\mathrm{tan}\:\:{x}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:{x}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{6}\:}]{\mathrm{1}+\mathrm{tan}\:{x}}\:−\mathrm{1}}\:? \\ $$ Answered by john santu last updated on 27/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\:\sqrt[{\mathrm{8}\:}]{\mathrm{1}+\frac{\mathrm{tan}\:{x}}{\mathrm{256}}}−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:{x}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{6}\:}]{\mathrm{1}+\mathrm{tan}\:{x}}\:−\mathrm{1}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 115604 by bemath last updated on 27/Sep/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{tan}\:{x}}\:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}\:+\mathrm{cos}\:{x}}\:−\mathrm{1}}{\left(\pi−{x}\right)^{\mathrm{2}} }\:=\:? \\ $$ Commented by Dwaipayan Shikari last updated on 27/Sep/20…
Question Number 181104 by cortano1 last updated on 21/Nov/22 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}}\right)}{\pi−\mathrm{arctan}\:\left(\mathrm{2x}\right)}\:=?\: \\ $$ Commented by Frix last updated on 21/Nov/22 $$\mathrm{wrong}. \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{4}}{{x}}\right)}{\pi−\mathrm{arctan}\:\mathrm{2}{x}}\:=\frac{\mathrm{0}}{\frac{\pi}{\mathrm{2}}}=\mathrm{0} \\…
Question Number 115551 by bemath last updated on 26/Sep/20 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}\:=\:? \\ $$ Answered by TANMAY PANACEA last updated on 26/Sep/20…
Question Number 115541 by bobhans last updated on 26/Sep/20 $$\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\left(\mathrm{2}−\frac{{x}}{{a}}\right)^{\mathrm{tan}\:\left(\frac{\pi{x}}{\mathrm{2}{a}}\right)} =? \\ $$ Answered by bemath last updated on 26/Sep/20 $${L}\:=\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\left(\mathrm{2}−\frac{{x}}{{a}}\right)^{\mathrm{tan}\:\left(\frac{\pi{x}}{\mathrm{2}{a}}\right)} \\ $$$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow{a}}…
Question Number 49927 by Abdo msup. last updated on 12/Dec/18 $${find}\:{lim}_{{x}\rightarrow{a}^{+} } \:\:\:\:\left({x}−{a}\right)\left({a}^{{x}} −{x}^{{a}} \right)\:\:\:{with}\:{a}>\mathrm{0}\:. \\ $$ Commented by afachri last updated on 12/Dec/18 $$\mathrm{i}\:\mathrm{found}\:\mathrm{0}\:,\mathrm{Sir}.\:…
Question Number 49817 by Aditya789 last updated on 11/Dec/18 $$\mathrm{prove}\:\mathrm{that}.\:\frac{\mathrm{a}^{\mathrm{r}} −\mathrm{1}}{\mathrm{r}}=\mathrm{1} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18 $${what}\:{is}\:{the}\:{relation}\:{between}\:{a}\:{and}\:{r}…{question} \\ $$$${not}\:{clear}… \\ $$…
Question Number 49803 by maxmathsup by imad last updated on 10/Dec/18 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\left({sinx}\right)^{{x}} \:−\mathrm{1}}{{x}^{{sinx}} \:−\mathrm{1}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com