Does the limit exists here or not?

The limit exists

The numerator and denominator both go to 0 as x->0. If this is a calculus course, you can use L'Hospital's Rule and differentiate numerator and denominator.

If not, try rationalizing the denominator. Multiply numerator and denominator by sqrt(x + 1) + 1

Result:

Note that x^3 + x^2 = x^2(x + 1) so sqrt(x^3 + x^2) = x sqrt(x + 1)

In the numerator you get x sqrt(x + 1) [ sqrt(x + 1) + 1].

In the denominator you get [sqrt(x + 1) - 1] [sqrt(x + 1) + 1] = sqrt(x + 1)^2 - 1^2 = x + 1 - 1 = x

and the x's cancel out in numerator and denominator.

You're left with lim(x->0) sqrt(x + 1) [ sqrt(x + 1) + 1]