Question Number 135571 by bemath last updated on 14/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{8}}\:=? \\ $$ Commented by EDWIN88 last updated on 14/Mar/21 $$\mathrm{i}\:\mathrm{guess}\:\mathrm{the}\:\mathrm{series}\:\mathrm{should}\:\mathrm{be}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{k}}{\left(\mathrm{k}^{\mathrm{2}}…
Question Number 70025 by Tony Lin last updated on 30/Sep/19 $${use}\:\varepsilon-\delta\:{defintion}\:{to}\:{prove}\:{that} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{{x}}=\mathrm{1} \\ $$ Commented by mind is power last updated on 01/Oct/19…
Question Number 4448 by Rasheed Soomro last updated on 29/Jan/16 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{3}{e}^{{x}} −{e}^{−{x}} −\mathrm{2}}{{x}}=? \\ $$ Answered by Yozzii last updated on 29/Jan/16 $${One}\:{approach}\:{can}\:{involve}\:{the}\:{use}\:{of} \\…
Question Number 4281 by Filup last updated on 07/Jan/16 $$\mathrm{The}\:\mathrm{following}\:\mathrm{image}\:\mathrm{shows}\:\mathrm{the}\:\mathrm{functiond} \\ $$$${f}\left({x}\right)={xe}^{\frac{\mathrm{1}}{\mathrm{1}−{x}}} \:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:{g}\left({x}\right)={x}−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{as}\:\mathrm{to}\:\mathrm{why}\:\mathrm{as}\:\mid{f}\left({x}\right)\mid\rightarrow\infty, \\ $$$$\mathrm{that}\:{f}\left({x}\right)\rightarrow{g}\left({x}\right). \\ $$ Commented by Filup last…
Question Number 69710 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Commented by mathmax by abdo last updated on 27/Sep/19 $${we}\:{have}\:{x}^{\mathrm{2}} −\mathrm{6}{x}\:+\mathrm{5}\:={x}^{\mathrm{2}} −{x}\:−\mathrm{5}\left({x}−\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)−\mathrm{5}\left({x}−\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{5}\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}}…
Question Number 69709 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Commented by kaivan.ahmadi last updated on 26/Sep/19 $${lim}_{{x}\rightarrow\infty} \frac{\mathrm{2}{x}}{\mathrm{3}^{{x}} {ln}\mathrm{3}}={lim}_{{x}\rightarrow\infty} \frac{\mathrm{2}}{\mathrm{3}^{{x}} \left({ln}\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ Commented…
Question Number 135223 by benjo_mathlover last updated on 11/Mar/21 $${Limit}\: \\ $$$$\left({a}\right)\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)=? \\ $$$$\left({b}\right)\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{108}\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{1}\right)}=? \\ $$ Answered…
Question Number 69667 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by Kunal12588 last updated on 26/Sep/19 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {{lim}}\frac{{sec}\:{x}\:−\mathrm{2}}{\frac{\pi}{\mathrm{3}}−{x}} \\ $$$$=\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {{lim}}\frac{{sec}\:{x}\:{tan}\:{x}}{−\mathrm{1}} \\ $$$$=−\mathrm{2}×\sqrt{\mathrm{3}}=−\mathrm{2}\sqrt{\mathrm{3}} \\ $$…
Question Number 4132 by prakash jain last updated on 29/Dec/15 $$\mathrm{Is}\:\mathrm{there}\:{f}\left({n}\right)\:\mathrm{such}\:\mathrm{that} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:{f}\left({n}\right){x}^{{n}} \right]\neq\mathrm{0} \\ $$$${f}\left({n}\right)\:\mathrm{is}\:\mathrm{independent}\:\mathrm{of}\:{x}. \\ $$ Commented by Yozzii last…
Question Number 135190 by liberty last updated on 11/Mar/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{x}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\:\mathrm{dt}\:? \\ $$$$ \\ $$ Answered by metamorfose last updated…