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November 23, 2014

November 23, 2014

Posted by **Janet** on Sunday, February 13, 2011 at 10:14pm.

Can someone tell me if it's right or wrong? And if there's a different approach?

- Calculus -
**bobpursley**, Sunday, February 13, 2011 at 10:39pmyou are correct if that orig is correct.

are you certain the denominator is not

sqrt(x-2) That makes it a calculus problem.

- Calculus -
**Janet**, Sunday, February 13, 2011 at 10:50pmNo it only has the sqrt x. The -2 is outside of the sqrt.

Just so I know. Had it all been under the square root would I have multiplied the top and the bottom by that sqrt(x-2)?

- Calculus -
**Reiny**, Sunday, February 13, 2011 at 10:51pmI recall seeing this question posted several times and both MathMate and I both answered it.

I was going for a "search" to find solutions but was not successful.

I agree with bobpursley that there might be a typo here.

I used to teach basic limits this way:

Sub in the approach value into the expression

If you get:

1. a real value, then write down that value. You are done! next question!

2. if you get a/0, where a≠0, then the limit is unefined, or there is no limit

3. If you get 0/0, then you have a real Calculus limit question. Try factoring, rationalizing or some other fancy tricks you can think of.

in your case , sub in x = 2

expresssion = 2(0)/(√2 - 2) = 0/-.586 = 0

all done!

However, check your typing and make sure your brackets are in the right place.

- Calculus -
**Janet**, Sunday, February 13, 2011 at 10:55pmthe first 2 is centered on the division line which has x-2/(sqrt2)-2.

I think I placed the brackets right??

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