As a simple example, take the function ƒ(x) = x2 and let x0 = 2. In this case, limx→2 (x2) = 4. What this means that as x approaches 2, ƒ(x) gets arbitrarily close to the value 4. To see what is happening here, imagine the graph of this function (which is a parabola opened up and passing through the origin):
If you take your finger and trace the curve in that graph, as you finger gets closer and closer to 2 along the x axis, it gets closer and closer to 4 along the y axis.
In this particular case,ƒ(x0) = limx→x0 ƒ(x) (i.e., limx→2 (x2) = 4 = 22). This is not always the case.
To see an example where this is not true, consider the function ƒ1(x) = (x4 - 1)/(x - 1). This function is not defined for x = 1 (at x = 1, the denominator is zero). You can see this in the graph below, where a hole is in the plot of the function at x = 1:
Because this function is undefined at x = 1, in order to find the limit as x approaches 1, it is useful to calculate the function for values of x close to 1 on both sides and see if the numbers are trending to a specific value:
x | 0.75 | 0.9 | 0.99 | 0.999 | 1 | 1.001 | 1.01 | 1.1 | 1.25 |
ƒ1(x) | 2.734 | 3.439 | 3.940 | 3.994 | undefined | 4.006 | 4.060 | 4.641 | 5.766 |
As you can guess from the table above, as x approaches 1 from either side, ƒ1(x) gets arbitrarily close to 4. Thus,
A bit of algebraic manipulation reveals that the numerator of this function can be factored as into (x3 + x2 + x + 1)(x - 1). If we cancel x - 1 in the numerator with x - 1 in the denominator we get the function ƒ2(x) = x3 + x2 + x + 1. This function is identical to ƒ1(x) except at the point x = 1, as you can see in the graph below:
As you have no doubt noticed, ƒ2(1) = 4, and as you may have guessed, it is not a coincidence that limx→1 ƒ1(x) = ƒ2(1). We shall see the reason for this shortly.
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