00:01
Given the equation, fx, is equal to x plus 1 divided by x squared plus 3, you have to find out its limit at a, when x approaches infinity, and b, when x approaches minus infinity.
00:15
So first, let's just modify the equation in a more solvable form, that is, divide the numerator and denominator by x -square.
00:23
So that would be x divided by x square plus one divided by x squared divided by x squared plus three divided by x square so these will cancel out and we'll be left with something like this 1 divided by x plus 1 divided by x square divided by 1 plus 2 divided by x squared now we have the limit where x approaches infinity.
01:04
Just to find that out, let's just substitute at the value of x here, but infinity all of these.
01:11
So that would become something like 1 by infinity plus 1 by infinity squared, divided by 1 plus 3 by infinity squared...