Limit Calculation for Radical Functions

[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.

 

[tiny-tim]

Hi DAPOS!

(try using the X² button just above the Reply box )

… is it correct to square the top and then place the resulting polynomial under a square root (i.e. sqrt(x^2 + 3)^2)?
…
So, you'll have:
sqrt(lim x→∞ (x^2 + 3)/ (2x^4 + 5))
Divide above and below by x^4 and solve. Thanks.

(you meant (x² + 3)²/ (2x⁴ + 5) )

Yes, that's fine.

 

[SherlockOhms]

Thanks!

 

[dextercioby]

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

 

[SherlockOhms]

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

 

[tiny-tim]

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

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

 

[SherlockOhms]

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

 

[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.

 

[Mark44]

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.

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 → ∞.

 

[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!