Homework Help: 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.