How do you find the Limit of this? (L'Hopitals Rule)

[Rnotaria]

Limit √(4x^(2)+x) -2x
x→∞
(Should read like: Square root of (4x squared plus x) minus 2x

I know you have to use L'Hopital's rule and I multiplied it by the conjugate to get a denominator, but every time I apply the rule it just goes on an infinite loop. Does anyone know how to solve this please?

Answer is 1/4

 

[SteamKing]

I think you need to review when you can use L'Hopital and when you can't.

If your limit is truly what you have written, it doesn't exist.

 

[Rnotaria]

The answer is supposed to be 1/4 though. And the homework assignment is all about L'Hopital's Rule.

 

[LCKurtz]

I think SteamKing needs to adjust his sauna.

Show us what you got when you multiplied by the conjugate. At that point you should be able to evaluate it without LH's rule. Think about putting all the $x$'s in the denominator.

 

[Rnotaria]

After multiplying by the conjugate I got:

Lim x/[(4x^2+x)^.5 +2x]
x→∞

Which should read: x over the square root of (4x squared plus x) + 2x

Then when I apply the rule it just goes on an infinite loop where the square root of 4x squared plus x alternates between being on top and bottom

 

[Mark44]

After multiplying by the conjugate I got:

Lim x/[(4x^2+x)^.5 +2x]
x→∞

Which should read: x over the square root of (4x squared plus x) + 2x

Then when I apply the rule it just goes on an infinite loop where the square root of 4x squared plus x alternates between being on top and bottom

Factor x out of the numerator and denominator, and then you'll be able to take the limit without the use of L'Hopital's Rule.

You'll need to factor x² out of the terms in the radical so that it becomes x outside of the radical.

 

[Rnotaria]

Factor x out of the numerator and denominator, and then you'll be able to take the limit without the use of L'Hopital's Rule.

You'll need to factor x² out of the terms in the radical so that it becomes x outside of the radical.

Are you allowed to factor x out of something raised to a power though?

 

[Mark44]

Sure. For example, (a³b⁶)² = (a²b² * ab⁴)² = (a²b²)² * (ab⁴)².

 

[vanhees71]

I'd not try to use de L'Hospital's rule here but expand the root appropriately for x \gg 1, using
\sqrt{4 x^2+x}=2x \sqrt{1+\frac{1}{4 x}}.
Then it's pretty easy to find the limit.

 

[LCKurtz]

Show us what you got when you multiplied by the conjugate. At that point you should be able to evaluate it without LH's rule. Think about putting all the $x$'s in the denominator.

Please quote the post to which you are replying, which I assume is mine, quoted above.

After multiplying by the conjugate I got:

Lim x/[(4x^2+x)^.5 +2x]
x→∞

Which should read: x over the square root of (4x squared plus x) + 2x

Then when I apply the rule it just goes on an infinite loop where the square root of 4x squared plus x alternates between being on top and bottom

Did you not read my post? Why are you telling me again that LH rule doesn't work when I just suggested a way to do it without LH's rule?