Prove that `lim_(x->0) (f(x+h)+f(x-h)-2f(x))/h^2 = f''(x)`

If f(x),g(x)a n dh(x) are three polynomial of degree 2, then prove that phi(x)=|[f(x)g(x)h(x)];[f'(x)g'(x) h '(x)];[f' '(x)g' '(x )h ' '(x)]| is: constant

If f''(0)=k,kne0 , then the value of lim_(xto0)(2f(x)-3f(2x)+f(4x))/(x^2) is

Let f(x) be a twice-differentiable function and f"(0)=2. The evaluate: lim_(x->0)(2f(x)-3f(2x)+f(4x))/(x^2)

If f(x)=sgn(x)" and "g(x)=x^(3) ,then prove that lim_(xto0) f(x).g(x) exists though lim_(xto0) f(x) does not exist.

If f''(0)=k,kne0 then the value of lim_(xrarr0)(2f(x)-3f(2x)+f(4x))/(x^(2)) is

Instead of the usual definition of derivative Df(x), if we define a new kind of derivative D^*F(x) by the formula D*f(x)=lim_(h->0)(f^2(x+h)-f^2(x))/h ,w h e r ef^2(x) mean [f(x)]^2 and if f(x)=xlogx ,then D^*f(x)|_(x=e) has the value (A)e (B) 2e (c) 4e (d) none of these

If f(x) is differentiable at x=a then show that lim_(xrarr0)(x^2f(a)-a^2f(x))/(x-a)=2af(a)-a^2f^1(a)

If f(x) is differentiable and strictly increasing function, then the value of lim_(xto0)(f(x^2)-f(x))/(f(x)-f(0)) is

Let f(x) be a twice-differentiable function and f''(0)=2. Then evaluate lim_(xto0) (2f(x)-3f(2x)+f(4x))/(x^(2)).

If lim_(x->oo) f(x) exists and is finite and nonzero and if lim_(x->oo) {f(x)+(3f(x)−1)/(f^2(x))}=3 ,then find the value of lim_(x->oo) f(x)