# Homework Help: Another find the limit

1. Oct 17, 2004

### quasar987

I've tried multiplicating by the conjugate of the denominator and of the numerator but this leads to nothing I can see. How can this limit be evaluated? (the limit is to be taken when n goes to infinity)

$$\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}$$

Thanks for helping.

2. Oct 17, 2004

### Hurkyl

Staff Emeritus
Show us what you got when you did this.

3. Oct 17, 2004

### Leong

Divide the numerator and the denominator by $$\sqrt{n}$$.

4. Oct 17, 2004

### quasar987

Will do tomorrow. I gotta go to bed urgent.

If you mean "take $\sqrt{n}$ out of the num and denom", when you take the limit you get the undeterminate form 0/0. If that's not what you mean, I don't know what you mean.

5. Oct 18, 2004

### quasar987

Ok, so

$$\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}} \frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}} = \frac{\sqrt{n+7}\sqrt{n+2}+\sqrt{n+7}\sqrt{n}-\sqrt{n+5}\sqrt{n+2}-\sqrt{n+5}\sqrt{n}}{2}$$

which may be factorised into...

$$\frac{(\sqrt{n+7}-\sqrt{n+5})(\sqrt{n+2}+\sqrt{n})}{2}$$

If we multiply by the other conjugate, we get

$$\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}} \frac{\sqrt{n+7}+\sqrt{n+5}}{\sqrt{n+7}+\sqrt{n+5}} = \frac{12}{\sqrt{n+2}\sqrt{n+7}+\sqrt{n+2}\sqrt{n+5} -\sqrt{n+7}\sqrt{n}-\sqrt{n+5}\sqrt{n}}$$

which may be factorised into

$$\frac{12}{(\sqrt{n+2}-\sqrt{n})(\sqrt{n+5}+\sqrt{n+7})}$$

6. Oct 18, 2004

### arildno

You've made quite a few errors here
Multiply with 1*1 in this manner:
$$\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}=\frac{\sqrt{n+7}+\sqrt{n+5}}{\sqrt{n+7}+\sqrt{n+5}}*\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}*\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}$$
Hence, we get:
$$\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}=\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+7}+\sqrt{n+5}}$$

Last edited: Oct 18, 2004
7. Oct 18, 2004

### quasar987

And THEN...
:tongue:

Ok I get it now. Thanks everyone!