Question 1035838
a.  Suppose {{{x <> 0}}}.  Then {{{x^2}}} can be cancelled from top and bottom, and the numerator is a polynomial in which the leading term is {{{5x^3}}}, while the denominator is a polynomial with a leading term of {{{2x^3}}}.  Therefore as n goes to infinity, f(n) goes to 5/2.

b.  As n approaches -1, f(n) approaches -2/7.  (In fact f(x) is continuous at x = -1).

c. Returning to the argument used in (a), in any  neighborhood of x = 0, f(x) is defined except at x = 0 itself.  This means the limit, as n approaches 0, would be -4/5.  (There is a removable discontinuity at x = 0.)