Limits using series expansion

You can try this clever trick:

let x= 1/y

Then you will be evaluating the lim as y->0 of (1/y^2 + 3y)^1/2 -1/y

Should be pretty simple now I'd imagine...

Hi, I'm stuck on a question to do with limits and series expansion. (AQA FP3)
The question is
lim (x->infinity) ((x^2 + 3x)^(1/2) -x)

I considered the binomial expansion by taking out a factor of 3x to leave me with ((3x)^(1/2))((1/3)x +1)^(1/2)) -x
but when i expanded this there didn't seem to be anything conclusive. I'm assuming that x must become a denominator because the limit is infinity.
The answer in the back says 3/2

Any advice would be great thanks!

I don't understand how this helps to be honest. Plus it would become (1/y² +3/y)½ not just 3y

My bad it is 3/y; I was writing that at 3am .

Anyway:

lim y->0 (1/y^2 + 3/y)^1/2 -1/y

lim y->0 (1/y^2)^1/2(1 + 3y)^1/2 - 1/y

lim y->0 1/y(1 + 1/2(3y)+ (1/2)(-1/2)(3y)^2/2!+...) -1/y

lim y->0 1/y(1+ 3/2y -9/8y^2+...)-1/y

lim y->0 1/y + 3/2 -9/8y -1/y

lim y->0 3/2 -9/8y= 3/2.

The usual trick with this sort of limit is to multiply by 1, but I'm not sure if that's taught in FP3.

Would that do much?

It does for the right value of 1 i.e. A/A where A is suitably chosen...