1. Apr 28, 2013

### SherlockOhms

When calculating the limit of the function f(x) = (x^2 + 3)/ sqrt(2x^4 + 5) as x→∞, is it correct to square the top and then place the resulting polynomial under a square root (i.e. sqrt(x^2 + 3)^2)? Then you can rewrite the problem as the square root of the limit as x→∞ of the resulting function.
So, you'll have:
sqrt(lim x→∞ (x^2 + 3)/ (2x^4 + 5))
Divide above and below by x^4 and solve. Thanks.

2. Apr 28, 2013

### tiny-tim

Hi DAPOS!

(try using the X2 button just above the Reply box )
(you meant (x2 + 3)2/ (2x4 + 5) )

Yes, that's fine.

3. Apr 29, 2013

### SherlockOhms

Thanks!

4. Apr 29, 2013

### dextercioby

While for that particular limit it is correct, squaring a function to find a limit of it is generally wrong.

5. Apr 29, 2013

### SherlockOhms

What would be the correct way to approach taking the limit of a radical? One that holds up for the majority of limits.

6. Apr 29, 2013

### tiny-tim

but it works if the function is always non-negative?

7. Apr 29, 2013

### SherlockOhms

It works if it isn't approaching negative infinity, right?

8. Apr 29, 2013

### pwsnafu

What's going on is $\lim_{x\to\infty} f(x)^2 = L^2 \implies \lim_{x\to\infty}f(x) = \pm L$, so you need to go back and check what the sign of f(x) is for sufficiently large x.

9. Apr 29, 2013

### Staff: Mentor

Rather than square the numerator and denominator, the way to go would be to factor the expression inside the radical.

$$\frac{x^2 + 3}{\sqrt{2x^4 + 5}} = \frac{x^2(1 + 3/x^2)}{x^2\sqrt{2 + 5/x^4}}$$
$$= \frac{1 + 3/x^2}{\sqrt{2 + 5/x^4}}$$

Now take the limit as x → ∞.

10. Apr 30, 2013

### SherlockOhms

I had been taught that method but I was just curious whether the above would actually work consistently. I can see that it won't, especially with limits to - infinity. Thanks for all the help!