Question Number 69075 by Aditya789 last updated on 19/Sep/19 Commented by Prithwish sen last updated on 19/Sep/19 $$\mathrm{Divide}\:\mathrm{both}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator} \\ $$$$\mathrm{by}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1} \\ $$ Commented by…
Question Number 134549 by liberty last updated on 05/Mar/21 $$\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\left(\frac{{xe}^{{ax}} }{{a}−{b}}\:+\:\frac{{e}^{{bx}} }{\left({a}−{b}\right)^{\mathrm{2}} }\:−\:\frac{{e}^{{ax}} }{\left({a}−{b}\right)^{\mathrm{2}} }\right)\:=? \\ $$ Answered by EDWIN88 last updated on 05/Mar/21…
Question Number 3460 by prakash jain last updated on 14/Dec/15 $$\mathrm{If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} \:\mathrm{is}\:\mathrm{convergent}\:\mathrm{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{b}_{{n}} \:\mathrm{is}\:\mathrm{convergent}. \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sufficient}\:\mathrm{condition}\:\mathrm{so}\:\mathrm{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} ^{{n}} {b}_{{n}}…
Question Number 68966 by anaplak last updated on 17/Sep/19 Commented by kaivan.ahmadi last updated on 17/Sep/19 $${lim}_{{x}\rightarrow\frac{\pi^{+} }{\mathrm{2}}} \frac{\mathrm{1}}{{cosx}}=\frac{\mathrm{1}}{\mathrm{0}^{−} }=−\infty \\ $$ Commented by anaplak…
Question Number 68965 by anaplak last updated on 17/Sep/19 Commented by anaplak last updated on 17/Sep/19 $${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}}…