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posted by on .

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 - ,

    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 - ,

    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 - ,

    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

    all done!

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

  • Calculus - ,

    the 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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