# Simple limit question.

1. Jan 18, 2006

### Nothing000

I am trying to find the limit as x aproaches infinity to this expression:
sqrt(x)-sqrt(x^2 -1)
I am using the conjugate method by multiplying the expresion by [sqrt(x)-sqrt(x^2 -1)/sqrt(x)-sqrt(x^2 -1)], and then dividing each term by the highest degree in the denominator which is 1/2. I am coming up with the limit as 0/1, or simply zero. But my calculator is coming up with -infinity. What am I doing wrong?

2. Jan 18, 2006

### benorin

The conjugate would be of opposite sign, so use

[sqrt(x)+sqrt(x^2 -1)/sqrt(x)+sqrt(x^2 -1)]

3. Jan 18, 2006

### benorin

limit at infinity

$$\lim_{x\rightarrow\infty} \sqrt{x}-\sqrt{x^2 -1} = \lim_{x\rightarrow\infty} \left( \sqrt{x}-\sqrt{x^2 -1} \right) \frac{\sqrt{x}+\sqrt{x^2 -1}}{\sqrt{x}+\sqrt{x^2 -1}} = \lim_{x\rightarrow\infty} \frac{x-\left( x^2 -1\right)}{\sqrt{x}+\sqrt{x^2 -1}}$$

the highest power of x on both top and bottom is 1 (since for x>0 $$\sqrt{x^2}=x$$,) so

$$= \lim_{x\rightarrow\infty} \frac{x-x^2 +1}{\sqrt{x}+\sqrt{x^2 -1}}\left( \frac{\frac{1}{x}}{\frac{1}{x}}\right) = \lim_{x\rightarrow\infty} \frac{1-x +\frac{1}{x}}{\sqrt{\frac{x}{x^2}}+\sqrt{\frac{x^2 -1}{x^2}}} =\lim_{x\rightarrow\infty} \frac{1-x +\frac{1}{x}}{\sqrt{\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}} =\frac{1-\infty +0}{\sqrt{0}+\sqrt{1-0}} = -\infty$$

Last edited: Jan 18, 2006
4. Jan 18, 2006

### Nothing000

5. Jan 18, 2006

### Nothing000

I don't understand what you did benorin. I get you up to this point, but from here I thought that you had to divide everything by the largest power in the denominator, and that is 1/2.

6. Jan 18, 2006

### Nothing000

What do you mean that the highest power in the top and bottom is one? It looks like the highest power in the numerator is 2 to me. Do you mean that the highest power that is simultaniously in the top and bottom is one? If so, where do you see a power of one in the denominator. Am i supposed to ignor the square root and look at the value within the square root funcion or something like that?

7. Jan 18, 2006

### benorin

You're right, it's the highest power on the bottom: the highest power of x on bottom is 1, since $$\sqrt{x^2}=x$$, (for x>0 to avoid to absolute value) so divide numerator and denominator by x.

8. Jan 18, 2006

### Nothing000

Ah geez... It is always those stupid little things that get you. Thanks a bunch!