- #1

- 47

- 0

^{kx}+ 9)/(4

^{5x}+ 3) as x->inf, when K=5, K<5, K>5.

Answer: I divided each term by the highest power of 4, and was able to come up with 0 (K<5), 1 (K=5), inf (K>5). I'm pretty sure these are correct.

Was this the right method? Am I good to assume that dividing by the highest power of x is similar in some ways to dividing by the highest power of some common base, which is what I did here?

Let me know if I can explain myself better.