Evaluate `lim_(x to oo) (ax^(p) + bx^(p- 3) + c)/(a_(1)x^(q) + b_(1)x^(q-1) + C_(1)X^(q-3) + d_(1))`
Where `p gt 0, q gt 0`, a,b,c, `a_(1), b_(1),C_(1),d_(1)` are constants.

Explanation

Here three cases arise `p gt q` or `p = q` or `p lt q`
Case-I : `P gt q`
Case - I : `p gt q`
`underset(x to oo)(lim) (ax^(p) xx bx^(p-3) + c)/(a_(1)x^(q) + b_(1)x^(q-1) + c_(1)x^(q-3) + d_(1)) = underset(x to oo)(lim) (x^(p)(a + (b)/x^(3)) + (c )/(x^(p)))/(x^(q) (a_(1) (b_(1))/(x) = (c_(1))/(x^(3)) + (d_(1))/(x_(q))))`
`= underset(x to oo)(lim) x^((p - q)) [(a + (b)/(x^(3)) + (c )/(x^(p)))/(a_(1) + (b_(1))/(x) + (c_(1))/(x^(3)) + (d_(1))/(x^(q)))]`
Case-II `p lt q`
`underset(x to oo)(lim) (ax^(p) + bx^(p - 3) + c)/(a_(1) x^(q) + b_(1)x^(q - 1) + c_(1) x^(q-3) + d_(1)) = underset(x to oo)(lim) (a + (b)/(x^(3)) + (c )/(x^(p)))/(x^(q - p)(a_(1) + (b_(1))/(x) + (c_(1))/(x^(3)) + (d_(1))/(x^(q))) = 10`
Case-III p = q
`underset(x to oo)(lim) (ax^(p) + bx^(p -3) + c)/(a_(1) x^(q) + b_(1) x^(q -1) + c_(1) x^(q - 3) + d_(1)) = underset(x to oo)(lim) (a + (b)/(x^(3)) + (c )/(x^(p)))/(a_(1) + (b_(1))/(x) + (c_(1))/(x^(3)) + (d_(1))/(x^(q))) = (a)/(a_(1))`
Thus `underset(x to oo)(lim) (ax^(p) + bx^(p-3) + c)/(a_(1)x^(q) + b_(1)x^(q-1) + c_(1)x^(q-3) + d_(1)) = {{:(oo, p gt q),(0, p lt q),((a)/(a_(1)), p=q):}` where `p gt 0, q gt 0`