Lim of this function using L'Hopitals rule multiple times

[animboy]

Homework Statement

lim as x approached 2 from above of:

(x^1/2 - 2^1/2 + (x - 2)^1/2)/(x² - 4)^1/2

Homework Equations

L'Hopitals rule...

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.

 

[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).

 

[animboy]

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).

But won't that give us a different limit? since the function squared is essentially a different function to the original.

 

[yenchin]

But won't that give us a different limit? since the function squared is essentially a different function to the original.

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

 

[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.

 

[Hootenanny]

But won't that give us a different limit? since the function squared is essentially a different function to the original.

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!

 

[animboy]

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!

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?

 

[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.