Question Number 68965 by anaplak last updated on 17/Sep/19 Commented by anaplak last updated on 17/Sep/19 $${sir}\:{please}\:{give}\:{me}\:{the}\:{solution}\:{showing}\:{by}\:{taking}\:{right}\:{hand}\:{limit}\: \\ $$$${i}\:{mean}\:{to}\:{say}\:{in}\:{detailed} \\ $$ Commented by kaivan.ahmadi last…
Question Number 68885 by aliesam last updated on 16/Sep/19 $${if}\: \\ $$$${f}\left({x}\right)=\frac{\mid{x}\mid}{{x}} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$${find}\: \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\:\:{f}\left({g}\left({x}\right)\right) \\…
Question Number 3297 by prakash jain last updated on 09/Dec/15 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{convergent} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{ln}\:{n}\right)^{\mathrm{ln}\left(\mathrm{ln}\:{n}\right)} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 3298 by prakash jain last updated on 09/Dec/15 $$\mathrm{Prove} \\ $$$$\underset{{i}=\mathrm{2}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{i}^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by 123456 last updated on 10/Dec/15…
Question Number 3138 by prakash jain last updated on 05/Dec/15 $$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{that} \\ $$$$\mathrm{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{k}} }\:,\:{k}\in\mathbb{Z}^{+} ,\:{k}>\mathrm{1}\:\mathrm{is}\:\mathrm{irrational}. \\ $$ Commented by 123456 last updated on…
Question Number 134173 by mathlove last updated on 28/Feb/21 $$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{{x}}−\mathrm{2}}{\:\sqrt{{x}}−\mathrm{2}\sqrt{\mathrm{2}}}=? \\ $$ Answered by malwan last updated on 28/Feb/21 $$\underset{{x}\rightarrow\mathrm{8}} {{lim}}\:\frac{\:^{\mathrm{3}} \sqrt{{x}}−\mathrm{2}}{\:\sqrt{{x}}−\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }+\mathrm{2}\:^{\mathrm{3}}…
Question Number 134171 by liberty last updated on 28/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\left(\mathrm{px}\right)−\mathrm{cos}\:\left(\mathrm{px}\right)}\:=? \\ $$ Answered by malwan last updated on 28/Feb/21 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{1}+\left({x}−…\right)−\left(\mathrm{1}−…\right)}{\mathrm{1}+\left({px}−…\right)−\left(\mathrm{1}−…\right)}\:\:=\:\frac{\mathrm{1}}{{p}} \\ $$ Answered…
Question Number 68626 by lavi.kaur last updated on 14/Sep/19 $$\mathrm{l}{im}\:\:\frac{−{n}\left(\mathrm{2}−{a}\right)^{{n}} }{\left(\mathrm{2}−{a}\right)} \\ $$$${n}\rightarrow\infty \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68554 by Mikael last updated on 13/Sep/19 Commented by Prithwish sen last updated on 13/Sep/19 $$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{1}}{\mathrm{n}}\left[\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{0}}{\mathrm{n}}}+\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}\:+……+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{n}−\mathrm{1}}{\mathrm{n}}}\right] \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\underset{\boldsymbol{\mathrm{r}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}−\mathrm{1}} {\sum}}\:\:\boldsymbol{\mathrm{lim}}\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\:}\frac{\mathrm{1}}{\mathrm{1}+\frac{\boldsymbol{\mathrm{r}}}{\boldsymbol{\mathrm{n}}}}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}}…
Question Number 134079 by mnjuly1970 last updated on 27/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..{mathematical}\:\:\:\:{analysis}….. \\ $$$$\:\:\:\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\mathrm{1}:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\Gamma\left({x}\right)\right){cos}^{\mathrm{2}} \left(\pi{x}\right){dx}=\frac{{ln}\left(\mathrm{2}\pi\right)}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}\:..\checkmark \\ $$$$\:\:\:\:\:\:\mathrm{2}:\:\:{lim}_{{n}\rightarrow\infty} \frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left({n}+\mathrm{2}\right)}{\Gamma^{\mathrm{2}} \left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\mathrm{1}…\checkmark \\ $$ Answered by…