# Lim of this function using L'Hopitals rule multiple times

1. Sep 15, 2011

### animboy

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

lim as x approached 2 from above of:

(x1/2 - 21/2 + (x - 2)1/2)/(x2 - 4)1/2

2. Relevant equations

L'Hopitals rule...

3. The attempt at a solution

I tried susbtituting in 2 and found that I got 0/0 thus it was possible to use L'hopitals. I then got inf/inf, so I used it again, but now I keep getting inf/inf. So what do I do? I'm stumped. I tried factorising but the + and - signs prevented me.

Last edited: Sep 15, 2011
2. Sep 15, 2011

### yenchin

The reason you keep getting indeterminate form is probably because the pesky power of 1/2. Assuming the limit exists, instead of looking for the limit of f(x) where f(x) is your function there, consider instead looking for the limit of the *square* of f(x).

3. Sep 15, 2011

### animboy

But wont that give us a different limit? since the function squared is essentially a different function to the original.

4. Sep 15, 2011

### yenchin

Of course you would then need to take the square root at the end...

5. Sep 15, 2011

### yenchin

To be more explicit, if the limit of f and g exist, then lim(f g)=(lim f)(lim g). In particular, lim(f^2)=(limf)^2.

6. Sep 15, 2011

### Hootenanny

Staff Emeritus
It follows from the fact that for any two functions $f(x)$ and $g(x)$ such that the limits

$$\lim_{x\to x_0}f(x)\text{ and } \lim_{x\to x_0}g(x)$$

exist, then

$$\lim_{x\to x_0}[f(x) g(x)] = [\lim_{x\to x_0}f(x)][\lim_{x\to x_0}g(x)]$$

Edit: Seems that I was beaten to it!

7. Sep 15, 2011

### animboy

I tried this and the resulting functions again gave me 0/0 so I again used L'hopitals and after length algebra this is the expression I got:

{2 - (2/x)0.5 + [2(2(x3 - 3x2 + 3x - 2))0.5 - (x2 - 2x)0.5]/ [2x3 - 8x2 + 8x]}

What do I do now, substituting in 2 gives <1 + inf> where inf comes from the third expression in the equation which gives "(0 - 0)/0". SO what now?

8. Sep 15, 2011

### yenchin

Now check again. The square of your function should give something nice. You only need to apply L'Hopital *once* to get the answer. In particular your denominator x^2-4 now just becomes 2x. So when x goes to 2, it does not give trouble.