# Problem using L'Hospiital's Rule (indeterminate form: INF - INF)

1. Sep 19, 2009

### SpicyPepper

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

$$\stackrel{lim}{x\rightarrow\infty}(\sqrt{x^2+x} - x)$$

I have no idea how to do this. In my book, it says I want to convert $$\infty - \infty$$ forms into a quotient by getting a common denominator, rationalization or by factoring.

2. Sep 19, 2009

### Dick

I would say you want to multiply by (sqrt(x^2+x)+x)/(sqrt(x^2+x)+x). I.e. rationalize.

3. Sep 19, 2009

### Hurkyl

Staff Emeritus
Many, many things would work, really. Multiplying and dividing by 1/x, for example.

Why 1/x? Well, it might be more clear if you think of it as "factoring out an x" -- then you just move the x to be a 1/x in the denominator so that you can use L'hôpitals.

4. Sep 19, 2009

### tnutty

y = sqrt(x^2 + x) - x

= sqrt(x ( x + 1) ) - x

= sqrt(x) * sqrt(x+1) - x

The use the product rule for term 1, combined with chain rule.

or
y = sqrt(x^2 + x) - x

= 1/x(sqrt(x^2 + x) - x)
= 1/x *sqrt(x^2 + x) - 1
= sqrt( (1/x)^2 ) * sqrt ( x^2 + x) - 1
= sqrt( (1/x)^2 ( x^2 + x) ) - 1
= sqrt( 1 + 1/x ) - 1

the use l'hopital rule.

5. Sep 19, 2009

### Gregg

^do that

$$(\sqrt{x^2+x}-x).\frac{(\sqrt{x^2+x}+x)}{(\sqrt{x^2+x}+x)} = \frac{1}{\frac{x}{x}\sqrt{1+\frac{1}{x^2}}+1}$$

Last edited: Sep 19, 2009
6. Sep 19, 2009

### SpicyPepper

I tried out all your suggestions. This is really cool. Thx everyone.