We are given
f ( x ) = lim r → x { 2 r 2 [ ( f ( r ) ) 2 − f ( x ) f ( r ) ] r 2 − x 2 − r 3 e f ( r ) r } f(x)=\sqrt{\lim_{r\to x}\left\{\frac{2r^2\big[(f(r))^2-f(x)f(r)\big]}{r^2-x^2}-r^3e^{\frac{f(r)}{r}}\right\}} f ( x ) = r → x lim { r 2 − x 2 2 r 2 [ ( f ( r ) ) 2 − f ( x ) f ( r ) ] − r 3 e r f ( r ) } with f f f ( − ∞ , 0 ) ∪ ( 0 , ∞ ) (-\infty,0)\cup(0,\infty) ( − ∞ , 0 ) ∪ ( 0 , ∞ ) f ( 1 ) = 1 f(1)=1 f ( 1 ) = 1
We need a a a f ( a ) = 0 f(a)=0 f ( a ) = 0 e a e^a e a
Simplify the limit expression.
Inside the limit,
( f ( r ) ) 2 − f ( x ) f ( r ) = f ( r ) ( f ( r ) − f ( x ) ) . (f(r))^2-f(x)f(r)=f(r)\big(f(r)-f(x)\big). ( f ( r ) ) 2 − f ( x ) f ( r ) = f ( r ) ( f ( r ) − f ( x ) ) . Also,
r 2 − x 2 = ( r − x ) ( r + x ) . r^2-x^2=(r-x)(r+x). r 2 − x 2 = ( r − x ) ( r + x ) . Hence
2 r 2 [ ( f ( r ) ) 2 − f ( x ) f ( r ) ] r 2 − x 2 = 2 r 2 f ( r ) ( f ( r ) − f ( x ) ) ( r − x ) ( r + x ) . \frac{2r^2[(f(r))^2-f(x)f(r)]}{r^2-x^2}
=\frac{2r^2 f(r)(f(r)-f(x))}{(r-x)(r+x)}. r 2 − x 2 2 r 2 [( f ( r ) ) 2 − f ( x ) f ( r )] = ( r − x ) ( r + x ) 2 r 2 f ( r ) ( f ( r ) − f ( x )) . As r → x r\to x r → x
f ( r ) − f ( x ) r − x → f ′ ( x ) , r → x , r + x → 2 x , r 2 → x 2 , f ( r ) → f ( x ) . \frac{f(r)-f(x)}{r-x}\to f'(x),\qquad r\to x,\quad r+x\to 2x,\quad r^2\to x^2,\quad f(r)\to f(x). r − x f ( r ) − f ( x ) → f ′ ( x ) , r → x , r + x → 2 x , r 2 → x 2 , f ( r ) → f ( x ) . Therefore,
lim r → x 2 r 2 [ ( f ( r ) ) 2 − f ( x ) f ( r ) ] r 2 − x 2 = 2 x 2 f ( x ) f ′ ( x ) 2 x = x f ( x ) f ′ ( x ) . \lim_{r\to x}\frac{2r^2[(f(r))^2-f(x)f(r)]}{r^2-x^2}
=\frac{2x^2 f(x) f'(x)}{2x}=x f(x) f'(x). r → x lim r 2 − x 2 2 r 2 [( f ( r ) ) 2 − f ( x ) f ( r )] = 2 x 2 x 2 f ( x ) f ′ ( x ) = x f ( x ) f ′ ( x ) . And clearly,
lim r → x r 3 e f ( r ) / r = x 3 e f ( x ) / x . \lim_{r\to x} r^3 e^{f(r)/r}=x^3 e^{f(x)/x}. r → x lim r 3 e f ( r ) / r = x 3 e f ( x ) / x . So the given relation becomes
f ( x ) = x f ( x ) f ′ ( x ) − x 3 e f ( x ) / x . f(x)=\sqrt{x f(x)f'(x)-x^3e^{f(x)/x}}. f ( x ) = x f ( x ) f ′ ( x ) − x 3 e f ( x ) / x . Squaring,
( f ( x ) ) 2 = x f ( x ) f ′ ( x ) − x 3 e f ( x ) / x . (f(x))^2=x f(x)f'(x)-x^3e^{f(x)/x}. ( f ( x ) ) 2 = x f ( x ) f ′ ( x ) − x 3 e f ( x ) / x . Thus
x f f ′ = f 2 + x 3 e f / x . x f f' = f^2 + x^3 e^{f/x}. x f f ′ = f 2 + x 3 e f / x .
Use the substitution
y = f ( x ) x ⇒ f ( x ) = x y . y=\frac{f(x)}{x} \quad \Rightarrow \quad f(x)=xy. y = x f ( x ) ⇒ f ( x ) = x y . Then
f ′ ( x ) = y + x y ′ . f'(x)=y+xy'. f ′ ( x ) = y + x y ′ . Substitute into
x f f ′ = f 2 + x 3 e f / x . x f f' = f^2 + x^3 e^{f/x}. x f f ′ = f 2 + x 3 e f / x . We get
x ( x y ) ( y + x y ′ ) = ( x y ) 2 + x 3 e y . x(xy)(y+xy')=(xy)^2+x^3e^y. x ( x y ) ( y + x y ′ ) = ( x y ) 2 + x 3 e y . That is,
x 2 y ( y + x y ′ ) = x 2 y 2 + x 3 e y . x^2y(y+xy')=x^2y^2+x^3e^y. x 2 y ( y + x y ′ ) = x 2 y 2 + x 3 e y . Expanding left side,
x 2 y 2 + x 3 y y ′ = x 2 y 2 + x 3 e y . x^2y^2+x^3yy'=x^2y^2+x^3e^y. x 2 y 2 + x 3 y y ′ = x 2 y 2 + x 3 e y . Cancel x 2 y 2 x^2y^2 x 2 y 2
x 3 y y ′ = x 3 e y . x^3yy'=x^3e^y. x 3 y y ′ = x 3 e y . Since x ≠ 0 x\neq 0 x = 0 x 3 x^3 x 3
y y ′ = e y . yy'=e^y. y y ′ = e y . So
y ′ e − y = 1 y . y' e^{-y}=\frac{1}{y}. y ′ e − y = y 1 . A cleaner way is to write
y d y d x = e y ⇒ y e − y d y = d x . y\frac{dy}{dx}=e^y
\quad\Rightarrow\quad
y e^{-y}\,dy=dx. y d x d y = e y ⇒ y e − y d y = d x . Integrate:
∫ y e − y d y = ∫ d x . \int y e^{-y}\,dy=\int dx. ∫ y e − y d y = ∫ d x . Now
∫ y e − y d y = − ( y + 1 ) e − y + C . \int y e^{-y}\,dy=-(y+1)e^{-y}+C. ∫ y e − y d y = − ( y + 1 ) e − y + C . Hence
− ( y + 1 ) e − y = x + C . -(y+1)e^{-y}=x+C. − ( y + 1 ) e − y = x + C . Or
( y + 1 ) e − y = C − x . (y+1)e^{-y}=C-x. ( y + 1 ) e − y = C − x .
Use the initial condition f ( 1 ) = 1 f(1)=1 f ( 1 ) = 1
At x = 1 x=1 x = 1
y = f ( 1 ) 1 = 1. y=\frac{f(1)}{1}=1. y = 1 f ( 1 ) = 1. So
( 1 + 1 ) e − 1 = C − 1 ⇒ 2 e = C − 1. (1+1)e^{-1}=C-1
\quad\Rightarrow\quad
\frac{2}{e}=C-1. ( 1 + 1 ) e − 1 = C − 1 ⇒ e 2 = C − 1. Thus
C = 1 + 2 e . C=1+\frac{2}{e}. C = 1 + e 2 . Therefore,
( y + 1 ) e − y = 1 + 2 e − x . (y+1)e^{-y}=1+\frac{2}{e}-x. ( y + 1 ) e − y = 1 + e 2 − x .
Find a a a f ( a ) = 0 f(a)=0 f ( a ) = 0
Since
f ( a ) = a y ( a ) , f(a)=a\,y(a), f ( a ) = a y ( a ) , and a ≠ 0 a\neq 0 a = 0 0 0 0
y ( a ) = 0. y(a)=0. y ( a ) = 0. Substitute y = 0 y=0 y = 0
( 0 + 1 ) e 0 = 1 + 2 e − a . (0+1)e^0=1+\frac{2}{e}-a. ( 0 + 1 ) e 0 = 1 + e 2 − a . So
1 = 1 + 2 e − a . 1=1+\frac{2}{e}-a. 1 = 1 + e 2 − a . Hence
a = 2 e . a=\frac{2}{e}. a = e 2 . Therefore,
e a = e 2 / e . e^a=e^{2/e}. e a = e 2/ e .
Comparison with stored answer.
Our derived value is
e a = e 2 / e , e^a=e^{2/e}, e a = e 2/ e , not 2 2 2
So the stored correct answer appears inconsistent with the equation as stated.