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If-f-1-1-and-f-x-1-x-2-f-x-2-then-lim-x-f-x-




Question Number 264 by novrya last updated on 17/Dec/14
If f(1)=1 and f ′(x)=(1/(x^2 +(f(x))^2 )) then   lim_(x→∞)  f(x)= ....
Iff(1)=1andf(x)=1x2+(f(x))2thenlimxf(x)=.
Commented by 123456 last updated on 17/Dec/14
(dy/dx)=(1/(x^2 +y^2 ))  −dx+(x^2 +y^2 )dy=0,y(1)=1  M=−1⇒M_y =0  N=x^2 +y^2 ⇒N_x =2x  −μdx+μ(x^2 +y^2 )dy=0  M=−μ⇒M_y =−μ_y   N=μ(x^2 +y^2 )⇒N_x =μ_x (x^2 +y^2 )+2xμ
dydx=1x2+y2dx+(x2+y2)dy=0,y(1)=1M=1My=0N=x2+y2Nx=2xμdx+μ(x2+y2)dy=0M=μMy=μyN=μ(x2+y2)Nx=μx(x2+y2)+2xμ
Commented by 123456 last updated on 21/Dec/14
the equation is hard and its possible that cannot be doing using elementary functions...  i thinks thats  exist other ways to compute it...
theequationishardanditspossiblethatcannotbedoingusingelementaryfunctionsithinksthatsexistotherwaystocomputeit
Commented by 123456 last updated on 17/Feb/15
∫_1 ^x f′(t)dt=∫_1 ^x (dt/(t^2 +[f(t)]^2 ))≤∫_1 ^x (dt/(t^2 +1))  f(x)−1≤(π/2)−(π/4)=(π/4)  f(x)≤1+(π/4)  f′=(1/(x^2 +f^2 ))>0,x≥1  (1/(x^2 +f^2 ))→0,x→+∞
x1f(t)dt=x1dtt2+[f(t)]2x1dtt2+1f(x)1π2π4=π4f(x)1+π4f=1x2+f2>0,x11x2+f20,x+