This is because when x→∞, 7/x→ 0, 1/x → 0, (x-5)/3→∞.
Or: x→∞lim[(x? -5x+7)/(3x+ 1)]= x→∞lim[( 1-5/x+7/x？ )/(3/x+ 1/x？ )]=∞
This is because when x→∞,-5/x→ 0,7/x? →0，3/x→0， 1/x？ →0，3/x+ 1/x？ →0, there is a 1 on the molecule, and the denominator →0.
So integral fraction →∞.
Generally speaking,
x→∞lim[(a？ x^n+a？ x^(n- 1)+a？ x^(n-2) +....+a？ n？ )/(b？ x^m+b？ x^(m- 1)+b？ x^(m-3)+...+b？ m？ )]
When n >; This limit =∞ is in m; When n=m, this limit =a? /b？ ; When n