Chapter 2 | Limits
165
Example 2.17 Evaluating a Limit by Factoring and Canceling
x 2 −3 x 2 x 2 −5 x −3 .
Evaluate lim x →3
Solution Step 1. The function f ( x ) =
x 2 −3 x 2 x 2 −5 x −3
is undefined for x =3. In fact, if we substitute 3 into the function
we get 0/0, which is undefined. Factoring and canceling is a good strategy: lim x →3 x 2 −3 x 2 x 2 −5 x −3 = lim x →3 x ( x −3) ( x −3)(2 x +1) Step 2. For all x ≠3, x 2 −3 x 2 x 2 −5 x −3 = x 2 x +1 . Therefore, lim x →3 x ( x −3) ( x −3)(2 x +1) = lim x →3 x 2 x +1 . Step 3. Evaluate using the limit laws: lim x →3 x 2 x +1 = 3 7 .
x 2 +4 x +3 x 2 −9 .
2.13
Evaluate lim
x →−3
Example 2.18 Evaluating a Limit by Multiplying by a Conjugate
x +2−1
Evaluate lim
x +1 .
x →−1
Solution Step 1. x +2−1 x +1
has the form 0/0 at −1. Let’s begin by multiplying by x +2+1, the conjugate of
x +2−1, on the numerator and denominator: lim x →−1 x +2−1 x +2+1 x +2+1 . Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the ( x +1) in the denominator cancels out in the end: = lim x →−1 x +1 ( x +1) ⎛ ⎝ x +2+1 x +1 = lim x →−1 x +2−1 x +1 · ⎞ ⎠ .
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