# Help with a limit

1. Aug 27, 2005

### sony

Hi,

Lim(t->0) t / sqrt(4+t)-sqrt(4-t)

I've stared at this for like half an hour :( Could someone please give me some hints of how I start factorizing this? Thanks.

2. Aug 27, 2005

### MalleusScientiarum

Can you use L'Hopital's Rule?

3. Aug 27, 2005

### sony

No, sorry. We havnent gotten that far yet :)

4. Aug 27, 2005

### TD

Multiply nominator & denominator with the complement of the denominator, i.e. $\left(\sqrt {4 + t} + \sqrt {4 - t}\right)$

In the denominator, use $\left( {a - b} \right)\left( {a + b} \right) = a^2 - b^2$

Now, you should be able to cancel out a t and just fill in t = 0.

5. Aug 27, 2005

### Curious3141

The easiest way, use the binomial expansion (the general form for nonintegral exponents).

$$(1 + x)^{\frac{1}{2}} \approx 1 + \frac{1}{2}x$$ for $|x| << 1$

So

$$\lim_{t -> 0}\frac{t}{\sqrt{4 + t} - \sqrt{4 - t}} = \lim_{t -> 0}\frac{t}{2(\sqrt{1 + \frac{t}{4}} - \sqrt{1 - \frac{t}{4}})} = \lim_{t -> 0}\frac{t}{2[(1 + \frac{t}{8}) - (1 - \frac{t}{8})]} = \lim_{t -> 0}\frac{t}{\frac{t}{2}} = 2$$

6. Aug 27, 2005

### sony

TD: Then i get ( sqrt(4+t)-sqrt(4-t)) / 2

But the answer is supposed to be 2. I'm still stuck :(

7. Aug 27, 2005

### TD

It should be a + between those square roots in the nominator

8. Aug 27, 2005

### sony

Curious: I've never seen that formula before. But thanks.

9. Aug 27, 2005

### sony

TD: Gaaah, stupid me. Well I got it now. Thanks!

10. Aug 27, 2005

### TD

No problem

This tactic is often used to get rid off square roots.