Let f:R→R be a real function. The function f is double differentiable. If there exists n∈N and p∈R such that limx→∞xn f(x) = p and there exists limx→∞xn+1 f(x), then
limx→∞xn+1 f' is equal to
-np
np
n2p
np2
We have limx→∞ xnfx=p⇒ limx→∞ xn+1fxx=pUsing L'Hospital rule, we get limx→∞ n+1xnfx+xn+1f'x1=p⇒ n+1p+ limx→∞ xn+1f'x=p⇒ limx→∞ xn+1 f'x=-npFurther limx→∞. xn+2f'xx=-npUsing L'Hospital rule, we get limx→∞ n+2xx+1f'x+xn+2f''x1=-np⇒ -npn+2+ limx→∞ xn+2f''x=-np⇒ limx→∞ xn+2f'' x =np1+n
limx→∞xn+1 f'' (x) is equal to
np(1-n)
-np(1+n)
np(1+n)