Chapter 2 | Limits
167
1 x +2 +1
2.15
Evaluate lim
x +3 .
x →−3
Example 2.20 does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques. Example 2.20 Evaluating a Limit When the Limit Laws Do Not Apply
Evaluate lim x →0 ⎛ ⎝ 1
⎞ ⎠ .
x + 5 x
( x −5)
Solution Both 1/ x and 5/ x ( x −5) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that 1 x + 5 x ( x −5) = x −5+5 x ( x −5) = x x ( x −5) . Thus, lim x →0 ⎛ ⎝ 1 x + 5 x ( x −5) ⎞ ⎠ = lim x →0 x x ( x −5) = lim x →0 1 x −5 = − 1 5 .
Evaluate lim x →3 ⎛
⎞ ⎠ .
2.16
⎝ 1
4 x 2 −2 x −3
x −3 −
Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form lim x → a − h ( x ), we require the function h ( x ) to be defined over an open interval of the form ( b , a ); for a limit of the form lim x → a + h ( x ), we require the function h ( x ) to be defined over an open interval of the form ( a , c ). Example 2.21 illustrates this point.
Example 2.21 Evaluating a One-Sided Limit Using the Limit Laws
Evaluate each of the following limits, if possible. a. lim x →3 − x −3
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