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Question Number 1676 by hhhggvghhh last updated on 31/Aug/15

hhjghhjjgkggjggigfhpppkknbgjffffg  biuggjnggtbnkoyfhjiuhbhuklphh  jinbh678656kkiohhuggjngffjkkdooxk  ikhhkjhv2[65(√(√(58 889<lkkjkbmbbnnb)))  jkkhjm(√(nlm6jikmbh(√(jlbj))))  jljbnkbbnmnmmnjknbjnrurubvm_(nkjnbmm)   jjnbbnjnnj^   nkn    mkknjlmmKihkhjZzzθn5ψn6

$$\boldsymbol{{hhjghhjjgkggjggigfhpppkknbgjffffg}} \\ $$ $$\boldsymbol{{biuggjnggtbnkoyfhjiuhbhuklphh}} \\ $$ $$\boldsymbol{{jinbh}}\mathrm{678656}\boldsymbol{{kkiohhuggjngffjkkdooxk}} \\ $$ $$\boldsymbol{{ikhhkjhv}}\mathrm{2}\left[\mathrm{65}\sqrt{\sqrt{\mathrm{58}\:\mathrm{889}<\boldsymbol{{lkkjkbmbbnnb}}}}\right. \\ $$ $$\boldsymbol{{jkkhjm}}\sqrt{\boldsymbol{{nlm}}\mathrm{6}\boldsymbol{{jikmbh}}\sqrt{\boldsymbol{{jlbj}}}} \\ $$ $$\boldsymbol{{jljbnkbbnmnmmnjknbjnrurubv}}\underset{\boldsymbol{{nkjnbmm}}} {\boldsymbol{{m}}} \\ $$ $$\boldsymbol{{jjnbbnjnnj}}^{} \\ $$ $$\boldsymbol{{nkn}} \\ $$ $$ \\ $$ $$\boldsymbol{{mkknjlmmK}}{ihkhjZzz}\theta{n}\mathrm{5}\psi{n}\mathrm{6} \\ $$

Commented by123456 last updated on 31/Aug/15

XD

$$\mathrm{XD} \\ $$

Commented byRasheed Ahmad last updated on 01/Sep/15

Neither question nor your comment  I can understand! What XD is  meant?

$${Neither}\:{question}\:{nor}\:{your}\:{comment} \\ $$ $${I}\:{can}\:{understand}!\:{What}\:\mathrm{XD}\:{is} \\ $$ $${meant}? \\ $$

Commented by123456 last updated on 01/Sep/15

a happy face

$$\mathrm{a}\:\mathrm{happy}\:\mathrm{face} \\ $$

Answered by 123456 last updated on 31/Aug/15

lim_(x→a^+ ) f(x)=L  ∀ε>0,∃δ,0<x−a<δ,∣f(x)−L∣<ε

$$\underset{{x}\rightarrow{a}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\mathrm{L} \\ $$ $$\forall\epsilon>\mathrm{0},\exists\delta,\mathrm{0}<{x}−{a}<\delta,\mid{f}\left({x}\right)−\mathrm{L}\mid<\epsilon \\ $$

Answered by Yozzy last updated on 31/Aug/15

∫...∫∫∫x_1 x_2 ...x_n dx_1 dx_2 ...dx_n =(1/2^n )x_1 ^2 x_2 ^2 ...x_n ^2 +C=(1/2^n )Π_(r=1) ^n x_r ^2 +C

$$\int...\int\int\int{x}_{\mathrm{1}} {x}_{\mathrm{2}} ...{x}_{{n}} {dx}_{\mathrm{1}} {dx}_{\mathrm{2}} ...{dx}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{x}_{\mathrm{1}} ^{\mathrm{2}} {x}_{\mathrm{2}} ^{\mathrm{2}} ...{x}_{{n}} ^{\mathrm{2}} +{C}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\underset{{r}=\mathrm{1}} {\overset{{n}} {\prod}}{x}_{{r}} ^{\mathrm{2}} +{C} \\ $$

Answered by 123456 last updated on 31/Aug/15

lim_(x→0) (1+x)^(1/x) =e⇔lim_(x→±∞) (1+(1/x))^x =e  x→0^+ ⇒1/x→+∞  x→0^− ⇒1/x→−∞

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} ={e}\Leftrightarrow\underset{{x}\rightarrow\pm\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} ={e} \\ $$ $${x}\rightarrow\mathrm{0}^{+} \Rightarrow\mathrm{1}/{x}\rightarrow+\infty \\ $$ $${x}\rightarrow\mathrm{0}^{−} \Rightarrow\mathrm{1}/{x}\rightarrow−\infty \\ $$

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