limits involving infinity

[jeca86]

1)Find a formula for a function that has vertical asymptotes x=1 and x=3 and horizontal asymptote y=1.


2)Find the limit:

lim(as x approaches infinity) x+2/the square root of (9x^2 +1)

 

[Gene]

You want verticals at 1 & 3, so put
((x-1)(x-3)) in the denominator. The numerator can be any constant a.
You want it to equal 1 when x=>infty so just add 1 to the fraction, which =>0.
y=a/((x-1)(x-3))+1
should do it.

I'm going to assume you never heard of PEMDAS and meant
(x+2)/sqrt(9x^2+1)
Divide top & bottom by x to get
(1+2/x)/sqrt(9+1/x^2)
As x => infty that =>
1/sqrt(9) = 1/3