let ...
back substitute
take the limit now?
lim y-(4*y^1/2)+3/y^2-1
x-1
I multiplied the numerator and denomination by the conjugate:
y-(4*y^1/2)+3 *y+(4*y^1/2)-3
y^2-1 *y+(4*y^1/2)-3
In the numerator, I get
y^2-16y-9
denomination gives me
y^2-1(y+(4*y^1/2)-3
I can factor the denominator a bit and i'm left with
y^2-16y-9/(y-1)(y+1)*(y+(4*y^1/2)-3
I can see that i should have got a factor of (y-1) in the numerator so that I could cancel the (y-1 ) in the denominator. I'm not sure what I am doing wrong. What did I do wrong and what am I missing?
Let's see if we can sort out your question by converting to LaTeX...
I'm not sure if this is actually the equation you had in mind, or where the comes into play. However, if that is what you're trying to solve, then I can tell you that doing this:
...won't help one bit, because it doesn't equal what you seem to think it does. Rather:
And obviously this is unhelpful. Actually, you can do this problem without performing any operations. Consider the original limit:
You can see that the denominator is going to grow faster than the numerator, because it contains the highest power (2). So, as approaches infinity, the denominator is going to get larger and larger with respect to the numerator, which means the limit is zero. However, if you want to see the operations, just divide all the terms by to the highest power in any term (which, as before, is 2):
And then simplify: