Question Number 4589 by FilupSmith last updated on 09/Feb/16 $${a}_{{n}} =\sqrt{\mathrm{2}+{a}_{{n}−\mathrm{1}} } \\ $$$${a}_{\mathrm{1}} =\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}} \\ $$$${L}=? \\ $$…
Question Number 4587 by FilupSmith last updated on 09/Feb/16 $${L}=\underset{{i}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} {i}}{{i}+\mathrm{1}} \\ $$$${L}=? \\ $$ Commented by Yozzii last updated on 14/Feb/16 $${L}\:{does}\:{not}\:{exist}.\:{While}\:\mid{L}\mid=\underset{{i}\rightarrow\infty} {\mathrm{lim}}\frac{{i}}{{i}+\mathrm{1}}…
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Question Number 4500 by FilupSmith last updated on 02/Feb/16 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}!}{{n}^{{n}} }={L} \\ $$$$ \\ $$$$\mathrm{Solve}\:\mathrm{for}\:{L} \\ $$ Commented by Yozzii last updated on 02/Feb/16…
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…