For this problem, we're literally just going to put five in every time we see our X. So if we think about actually might reduce first, we don't have to, but we could two X + 5, but that denominator could have an X that factors out. If we factor that X out, then we can see that there's an X in this numerator and an X in that denominator. That could reduce. So we could get the limit as X approaches 5 from the right hand side of 7 / X + 1 times two X + 5 / X + 1. Now when I stick in the 576 times 2 * 5 + 5 or 15 / 6 so seven 6 * 15615 six can reduce a three will come out of each. So seven 6 * 5 halves would give us 35 twelves.