Calculus Volume 1

204

Chapter 2 | Limits

lim x → a + f ( x ) = L if for every ε >0, there exists a δ >0 such that if 0< x − a < δ , then | f ( x )− L | < ε . Limit from the Left: Let f ( x ) be defined over an open interval of the form ( b , c ) where b < c . Then, lim x → a − f ( x ) = L if for every ε >0, there exists a δ >0 such that if − δ < x − a <0, then | f ( x )− L | < ε .

Example 2.44 Proving a Statement about a Limit From the Right

Prove that lim x →4 +

x −4=0.

Solution Let ε >0.

Choose δ = ε 2 . Since we ultimately want | x −4−0 | < ε , we manipulate this inequality to get x −4< ε or, equivalently, 0< x −4< ε 2 , making δ = ε 2 a clear choice. We may also determine δ geometrically, as shown in Figure 2.42 .

Figure 2.42 This graph shows how we find δ for the proof in Example 2.44 .

Assume 0< x −4< δ . Thus, 0< x −4< ε 2 . Hence, 0< x −4< ε . Finally, | x −4−0 | < ε . Therefore, lim x →4 + x −4=0.

2.30

Find δ corresponding to ε for a proof that lim x →1 −

1− x =0.

We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by

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