Question-23012

Question Number 23012 by A1B1C1D1 last updated on 25/Oct/17

Commented by ajfour last updated on 25/Oct/17

i had solved your previous question ∫_0 ^( 2) ∫_(y/2) ^( 1) e^x^2 dxdy =?

$${i}\:{had}\:{solved}\:{your}\:{previous}\: \\ $$$${question}\:\int_{\mathrm{0}} ^{\:\:\mathrm{2}} \int_{{y}/\mathrm{2}} ^{\:\:\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dxdy}\:=? \\ $$

Commented by A1B1C1D1 last updated on 25/Oct/17

$$\mathrm{Yes}! \\ $$

Answered by ajfour last updated on 25/Oct/17

=lim_(x→0) ((([1+2x+((4x^2 )/(2!))+((8x^3 )/(3!))+]..−[1−2x+((4x^2 )/(2!))−((8x^3 )/(3!))]−4x)/(x−x+(x^3 /(3!))−...))) =lim_(x→0) (((2×((8x^3 )/(3!))+...)/((x^3 /(3!))+....))) = 16 .

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\left[\mathrm{1}+\mathrm{2}{x}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}!}+\right]..−\left[\mathrm{1}−\mathrm{2}{x}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}!}\right]−\mathrm{4}{x}}{{x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}−…}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{2}×\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{3}!}+…}{\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+….}\right)\:=\:\mathrm{16}\:. \\ $$

Commented by A1B1C1D1 last updated on 26/Oct/17

$$\mathrm{Thank}\:\mathrm{you}. \\ $$

Commented by A1B1C1D1 last updated on 26/Oct/17

You did the expansion of the sine function in infinite series, right?

$$\mathrm{You}\:\mathrm{did}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{function}\:\mathrm{in}\:\mathrm{infinite}\:\mathrm{series},\:\mathrm{right}?\: \\ $$

Answered by FilupS last updated on 26/Oct/17

lim_(x→0) ((e^(2x) −e^(−2x) −4x)/(x−sin(x))) lim_(x→0) ((2(((e^(2x) −e^(−2x) )/2))−4x)/(x−sin(x))) u=2x lim_(u→0) ((2(((e^u −e^(−u) )/2))−2u)/((u/2)−sin((u/2)))) lim_(u→0) ((2sin(u)−2u)/((u/2)−sin((u/2))))=(0/0) L′Hopitals rule lim_(u→0) ((2cos(u)−2)/((1/2)+(1/2)cos((u/2)))) L′Hopitals rule lim_(u→0) ((−2sin(u))/(−(1/4)sin((u/2)))) =(2/(1/4)) =8???

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}{x}} −{e}^{−\mathrm{2}{x}} −\mathrm{4}{x}}{{x}−\mathrm{sin}\left({x}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left(\frac{{e}^{\mathrm{2}{x}} −{e}^{−\mathrm{2}{x}} }{\mathrm{2}}\right)−\mathrm{4}{x}}{{x}−\mathrm{sin}\left({x}\right)} \\ $$$${u}=\mathrm{2}{x} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left(\frac{{e}^{{u}} −{e}^{−{u}} }{\mathrm{2}}\right)−\mathrm{2}{u}}{\frac{{u}}{\mathrm{2}}−\mathrm{sin}\left(\frac{{u}}{\mathrm{2}}\right)} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\left({u}\right)−\mathrm{2}{u}}{\frac{{u}}{\mathrm{2}}−\mathrm{sin}\left(\frac{{u}}{\mathrm{2}}\right)}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$${L}'{H}\mathrm{opitals}\:\mathrm{rule} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2cos}\left({u}\right)−\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\frac{{u}}{\mathrm{2}}\right)} \\ $$$${L}'{H}\mathrm{opitals}\:\mathrm{rule} \\ $$$$\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\left({u}\right)}{−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\frac{{u}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}/\mathrm{4}} \\ $$$$=\mathrm{8}??? \\ $$

Commented by A1B1C1D1 last updated on 26/Oct/17