The problem is lim --> 2 for g(x) which is 2((x-2/(squareroot x)-2)) I just substituted 2 in for x and 2((2-2)/(squareroot 2)-2) = 0 This doesn't look right. It seems like I'd need to try something different.
Can someone tell me if it's right or wrong? And if there's a different approach?
Calculus - bobpursley, Sunday, February 13, 2011 at 10:39pm
you 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 10:50pm
No 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 10:51pm
I 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
However, check your typing and make sure your brackets are in the right place.
Calculus - Janet, Sunday, February 13, 2011 at 10:55pm
the first 2 is centered on the division line which has x-2/(sqrt2)-2.
I think I placed the brackets right??