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Sunday, February 12, 2006


So, in today's class, February 10th, we started unit one which's limits.
We looked at three different graphs and we find the limit for each graph.
We learned that when substitution is used to evaluate a limit, if the numerator
is not zero and the denominator is zero, then the limit does not exist.

When evaluation the limit by direct substitution results in zero in the numerator, as well as zero in the denominator, the resulting limit is said to be of an Indeterminate form. The value of such a limit, if it converges on some real value, is not always obvious. This form requires algebraic manipulation to evaluate the limit correctly.

For example if u are given a question that says find the limiting value, as x approaches 2.

f(x)=2x3 What you do is substitution. f(2)=2(2)+3=7, therefore the limit is =7.

The next scribe maker will be Calvinw

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