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Limits Part II

Updated on July 8, 2010

Limits Continued

This is a continuation of the previous lesson on limits.  If you would like to see the first lesson on limits, please visit http://hubpages.com/hub/limits
.

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Now, let us continue with our study of limits.  We ended with the fact that we must sometimes place a condition on the limit (is the limit coming from the right or left?).  Now, let us look at another example, in which there may arise some problems.

EXAMPLE I: Find the limit as x approaches -3 of the function (x2 -9)/(x+3).

/ is equivalent to divided by.  I am sorry about the lack of math symbols.

Now the first step always is to check if this limit may exist without modification.  To do so, simply plug in -3 into the x values:

(32 -9)/(-3+3)=(0/0) ???

Does this limit exist?  Of course it does!  Why?  Allow me to explain.

Remember, when we are mentioning limits, the limit does not go to the designated point, but approaches it.  What really is the function above?  It is a line with a hole at x=-3.  At first this may not be that apparent, but in order to actually find the value at which the hole exists, we must find a way to simplify the function.

(x2 -9)=(x+3)(x-3)

Therefore, we get ((x+3)(x-3))/(x+3)  We can take out the x+3 as they are common factors in both the numerator and denominator.  Our result is then x-3, which is the equation of the line.  Now all we have to do is plug in x, which is -3 into the equation (-3-3=-6).  The answer is -6.


Now, when we want the limit to approach infinity or negative infinity, then there are times when the limit may approach a finite number.  This will not happen with polynomials (all positive integer exponents), but when there is a denominator with the variable, then you may want to watch out.

EXAMPLE II: Find the limit as x approaches infinity of the function 1/x.

Now, we must use a concept called end behavior when dealing with limits of infinity.  Now, to enact this end behavior, we simply divide by the highest power of x present in the function to all the terms.

The highest powered term of x is x, therefore we must divide by x on both the numerator and denominator.

limx→ ∞  (1/x) / (x/x)

Now for the numerator (1/x), if the denominator were to become extremely large, the value of the denominator would approach zero.  Therefore, 1/x could be designated as 0 (as well as (x2 /x3, 5/x).  x/x is simply 1. 0/1 is zero, therefore the limit as x approaches infinity of (1/x) is 0.  If you graph this you will see that the function tries to reach 0, but is very close to it as x gets larger.  This limit to infinity helps find the horizontal asymptote (functions can only have one).

That concludes our lesson on limits Part II.  The next page will go more in-depth about limits of derivatives and trigonometric derivatives.

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