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), thenlimx→∞xn+1 f' is equal tolimx→∞xn+1 f'' (x) is equal to

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+nWe  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