Limits approaching infinity...

[sydney_bristow87]

Give me hope that I at least got one question right on my exam... :?

This is a relatively easy problem, but I have no idea how to solve it. I attempted to use L hopitals rule (which I had skimmed over the night before, and probably don't know how to use correctly, but at the time, I could think of no other way to solve it... I still can't, actually) and got 18/4 ... anyway, the problem:

what is the limit as x approaches infinity : (3x + 2) / the square root of (2x^2 + 3)

and the limit as x approaches negative infinity, same problem...

 

[Gene]

You don't need L'H for this. Just dividr numerator and denomonator by x.
(3x + 2) /sqrt(2x^2 + 3) =
(3 + 2/x) /sqrt(2 + 3/x²)
As x approaches infinity the x in the denominator causes those terms to vanish and you are left with
3/+sqrt(2). The tricky part is that in the original equation the denominator is always positive but if x is positive the limit is positive. If x is negative the limit is negative so to the right you use the +sqrt, to the left you use the -sqrt.