On-line Math 21

1.3.1  Examples of computing limits using rules and theorems.

Worked Examples

Example 1 Compute

lim x® 3  x2-x-6 x3-4x2+3x .

This is a typical example of how to deal with limits. There is a problem with that formula when x = 3 , in that both the numerator and the denominator are 0 there. In order to make sense of it, we need to see why those terms are 0 at x = 3 . But there is no mystery here, since both the numerator and the denominator can be factored,

x2-x-6 = (x-3)(x+2),

and

x3-4x2+3x = x(x-3)(x-1),

the expression

x2-x-6 x3-4x2+3x

can be simplified to

x2-x-6 x3-4x2+3x = (x-3)(x+2) x(x-3)(x-1) = (x+2) x(x-1) ,

so the limit of the original is the same as the limit of the simplified version,

lim x® 3  x2-x-6 x3-4x2+3x = lim x® 3  (x+2) x(x-1) = 3+2 3(3-1) = 5 6 .

At that last step we use the fact that, for a simple formula that has no ambiguity, you can find the limit just by plugging the value in. This is technically the fact that the function

g(x): = (x+2) x(x-1) ,

which equals f(x) except for the fact that g(x) is defined at x = 3 , is a continuous function.

Example 2 Compute

lim x® -1  Ö   x3+2x+7   .

To be fair, we really don't have the theory to claim this (yet), but we can argue simply that, since

lim x® -1  x3+2x+7 = lim x® -1  æ è Ö   x3+2x+7   ö ø 2   = lim x® -1  Ö   x3+2x+7   lim x® -1  Ö   x3+2x+7   ,

and

lim x® -1  x3+2x+7 = (-1)3+2(-1)+7 = 4,

then

lim x® -1  Ö   x3+2x+7   = 2.

Example 3 Compute

lim x® 9  x2-81 Öx-3 .

Example 4 Define

f(x): = ì ï í ï î   ___ Öx-4   , if x > 4 8-2x, if x < 4 .

Find

lim x® 4  f(x),

if it exists.

Example 5 Find

lim x® 0  xsin( 1 x )

by using the squeeze theorem.