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How can I find the lim of (3 - x^2)/(x-3) as x approaches to 3?
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We need to find the value of lim x-->3 [ (3 - x^2)/(x - 3)]
Substitute x = 3, we get -6/0
We see that the denominator has a root at x = 3. Therefore the graph has a vertical asymptote at x = 3.
Also, for values of x > 3, (3 - x^2)/(x - 3) is negative and for values of x less than 3, the graph is positive. At x = 3, it does not have any value.
The required limit is not defined and cannot be found.
Posted by justaguide on April 9, 2011 at 4:05 PM (Answer #1)
To check if the function has limit, when x approaches to 3, we'll determine the lateral limit and the value of the function in accumulation point.
For x->3, x<3 (x approaches to 3, with values smaller than 3)
lim (3 - x^2)/(x-3) = -6/-0 = +infinite
For x->3, x>3 (x approaches to 3, with values higher than 3)
lim (3 - x^2)/(x-3) = -6/+0 = -infinite
Since the lateral limits are infinite and different, the limit of the function does not exist if x approaches to 3.
Posted by giorgiana1976 on April 9, 2011 at 4:22 PM (Answer #2)
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