Calc Please Help

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Are these correct?

lim x->0 (x)/(sqrt(x^2+4) - 2)
I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity?

lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2)
I get 2/0, so lim x-> 1+ = - infinity? and lim x->1- = + infinity?

lim h->0 [(-7)/(2+h^2) + (7/4)]/h
I used a computational website to get (7/4) as the answer, but I did not get this. My work ends with: (-28)/(4(h+8) + (7(h+8)/(4(h+8)) and I end with -5.25 for an answer?

Last one!

lim x-> neg. infinity (-2x^2 + 3x - 2)/(5x^3 + 4x -x + 1)
Don't even know about this one?

Please help. I would like to understand these.

  • Calc Please Help -

    The first two are correct.

    Looking at the pattern of the third and considering the answer , I think you have a typo
    The function appears to be f(x) = -7/x^2 and you are finding the derivative when x = 2

    Instead of lim h->0 [(-7)/(2+h^2) + (7/4)]/h , it should be
    lim h->0 [(-7)/(2+h)^2 + (7/4)]/h , notice the change in brackets.

    then I get [-28 + 7(2+h)^2]/(4(2+h)^2) / h
    = [ -28 + 28 + 28h + 7h^2]/(4(2+h)^2) / h
    = h(28+7h)/(4(2+h)^2) / h
    = (28+7h)/(4(2+h)^2
    now as h ---> 0 this becomes
    28/16 = 7/4

    for the last one, how about dividing each term in both the numerator and denominator by the highest power of x that you see, that is , by x^3
    to get the expression as

    (-2/x + 3/x^2 - 2/x^3)/(5 + 4/x^2 - 1/x^2 + 1/x^3)

    Now consider each term
    As x becomes hugely negative, each of the terms would still be negative, but very very close to zero, so the top approaches zero, the bottom obvious approaches 5
    so you have -0/5 which approaches 0 from the negative.

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