# Homework Help: How would I solve this limit?

1. Dec 8, 2009

### holezch

1. The problem statement, all variables and given/known data

(2+y^2)/sqrt(4+y^4) as y goes to positive infinity

2. Relevant equations

3. The attempt at a solution

I divided it out by y^2 to get

lim y-> inf+ (2+y^2)/y^2 = 1 and the bottom lim y -> inf+ sqrt(4+y^4)/y^2 I can't solve..

by the way, I can't use L'hopitals rule

thanks!!

2. Dec 8, 2009

### Unto

Split them into 2.

1. $$2 + y^2$$ --> inf as y --> inf
2. $$1/sqrt(4 + y^4)$$ --> 0 as y -- inf

So you are having an inf / 0 situation.

3. Dec 8, 2009

### holezch

I know, that's why divided it out by y^2.. so the question really comes down to solving

lim y -> inf+ sqrt(4+y^4)/y^2

4. Dec 8, 2009

### grief

First of all, notice that (2+y^2)^2=(4+4y^2+y^4)>(4+y^4) for y positive.
This means (2+y^2)>sqrt(4+y^4)... which means that the limit you're looking for is either greater than or equal to 1 or infinite (why?).

As for evaluating the actual limit, you might find it easier to first deal with the ratio (4+4y^2+y^4)/(4+y^4)

5. Dec 8, 2009

### Staff: Mentor

Factor y^2 out of both the numerator and denominator and then take the limit.
$$\frac{2 + y^2}{\sqrt{4 + y^4}}~=~\frac{y^2(2/y^2 + 1)}{y^2\sqrt{4/y^4 + 1}}$$

Can you finish it?

6. Dec 8, 2009

### holezch

thanks, I can see that for large y, it is 1

7. Dec 8, 2009

### Staff: Mentor

Yes. And you can see that in the original expression, as well. For large y, the numerator is pretty close to y^2, and the denominator is also pretty close to y^2, since the square root of y^4 is y^2. The constants in the top and bottom are less and less significant the larger y gets.