Expert: Sherman D. - 5/4/2005

Question
My questions are the following:  Find the limit of each function as x approaches infinty and as x appraoches negative infinity.

1) g(x) = [1]/ 2+(1/x)

2) h(x) = [2-(2/x)]/4+(sqrt(2)/x^2)

3) h(x) = [-5=(7/x)]/3-(1/x^2)

If it helps at all, the back of the book says the answer to number 3 is     -5/3.  I know that when you are dealiing with limits and infinity that you are supposed to divide by the highest power x from the denominator. Let me know is any of this make sense.

thanks

Answer
1.) If by this you mean
g(x) = 1/(2 + (1/x))

2 + (1/x)
(2x + 1)/x

1/((2x + 1)/x))
(1/1)/((2x + 1)/x)
(1/1)*(x/(2x + 1))
x/(2x + 1)

ANS : (1/2)

www.calculator.com will show you that my answer is correct.

I wish i could mathematically proove it, but i am just going by what you told me about #3, and by using the graph site, and so far it seems to be working. If i had actual values instead of ±infinity, it would be easier to work with, because then i could plug in actuall values and then i could proove if the answer is correct mathematically.

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2.)
h(x) = ((2 - (2/x))/4 + (sqrt(2)/(x^2))
h(x) = ((2x - 2)/x)/4 + (sqrt(2)/(x^2))
h(x) = (((2x - 2)/x)*(1/4)) + (sqrt(2)/(x^2))
h(x) = ((2x - 2)/(4x)) + (sqrt(2)/(x^2))
multiply everything by 4x^2
h(x) = ((2x - 2)x) + 4sqrt(2))/(4x^2)
h(x) = (2x^2 - 2x + 4sqrt(2))/(4x^2)
h(x) = (2(x^2 - x + 2sqrt(2))/(2(2x^2))
h(x) = (x^2 - x + 2sqrt(2))/(2x^2)

following what you had on #3, if you take the coefficients of the highest "x" values,

ANS : (1/2)

If you check out the graphing site, it will show you the same thing.

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3.)
h(x) = (-5 + (7/x))/3 - (1/(x^2))
h(x) = ((-5x + 7)/x)/3 - (1/(x^2))
h(x) = (((-5x + 7)/x)/(3/1)) - (1/(x^2))
h(x) = (((-5x + 7)/x)*(1/3)) - (1/(x^2))
h(x) = ((-5x + 7)/(3x)) - (1/(x^2))
multiply everything by 3x^2
h(x) = (((-5x + 7)x) - 3x)/(3x^2)
h(x) = (-5x^2 + 7x - 3x)/(3x^2)
h(x) = (-5x^2 + 4x)/(3x^2)
h(x) = (x(-5x + 4))/(3x^2)
h(x) = (-5x + 4)/(3x)

At the graph site i told you about, it shows that the limit is (-5/3) as "x" approaches ±infinity.

ANS : (-5/3)

These only work because your approaching ±infinity.