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Chapter 2 | Limits
Step 3. Then we cancel:
1 x +2+1 .
= lim
x →−1
Step 4. Last, we apply the limit laws:
1 x +2+1
lim x →−1
= 1 2 .
2.14
x −1−2 x −5 .
Evaluate lim x →5
Example 2.19 Evaluating a Limit by Simplifying a Complex Fraction
1 x +1 −
1 2 x −1 .
Evaluate lim x →1
Solution
1 x +1 − x −1
1 2
has the form 0/0 at 1. We simplify the algebraic fraction by multiplying by
Step 1.
2( x +1)/2( x +1) :
lim x →1 2( x +1) 2( x +1) . Step2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor ( x −1): = lim x →1 2−( x +1) 2( x −1)( x +1) . Step 3. Then, we simplify the numerator: = lim x →1 − x +1 2( x −1)( x +1) . Step 4. Now we factor out −1 from the numerator: = lim x →1 −( x −1) 2( x −1)( x +1) . Step 5. Then, we cancel the common factors of ( x −1): = lim x →1 −1 2( x +1) . Step 6. Last, we evaluate using the limit laws: lim x →1 −1 2( x +1) = − 1 4 . 1 2 x −1 = lim x →1 1 x +1 − 1 2 x −1 · 1 x +1 −
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