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Proving a statement about the limit of a specific function (algebraic approach)

Prove that lim x −1 ( x 2 2 x + 3 ) = 6 .

Let’s use our outline from the Problem-Solving Strategy:

  1. Let ε > 0 .
  2. Choose δ = min { 1 , ε / 5 } . This choice of δ may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: | ( x 2 2 x + 3 ) 6 | < ε . This inequality is equivalent to | x + 1 | · | x 3 | < ε . At this point, the temptation simply to choose δ = ε x 3 is very strong. Unfortunately, our choice of δ must depend on ε only and no other variable. If we can replace | x 3 | by a numerical value, our problem can be resolved. This is the place where assuming δ 1 comes into play. The choice of δ 1 here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since δ 1 and | x + 1 | < δ 1 , we are able to show that | x 3 | < 5 . Consequently, | x + 1 | · | x 3 | < | x + 1 | · 5 . At this point we realize that we also need δ ε / 5 . Thus, we choose δ = min { 1 , ε / 5 } .
  3. Assume 0 < | x + 1 | < δ . Thus,
    | x + 1 | < 1 and | x + 1 | < ε 5 .

    Since | x + 1 | < 1 , we may conclude that −1 < x + 1 < 1 . Thus, by subtracting 4 from all parts of the inequality, we obtain −5 < x 3 < 1 . Consequently, | x 3 | < 5 . This gives us
    | ( x 2 2 x + 3 ) 6 | = | x + 1 | · | x 3 | < ε 5 · 5 = ε .

    Therefore,
    lim x −1 ( x 2 2 x + 3 ) = 6 .
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Complete the proof that lim x 1 x 2 = 1 .

Let ε > 0 ; choose δ = min { 1 , ε / 3 } ; assume 0 < | x 1 | < δ .

Since | x 1 | < 1 , we may conclude that −1 < x 1 < 1 . Thus, 1 < x + 1 < 3 . Hence, | x + 1 | < 3 .

| x 2 1 | = | x 1 | · | x + 1 | < ε / 3 · 3 = ε

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You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits.

Proving limit laws

We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality    is used at a key point of the proof, so we first review this key property of absolute value.

Definition

The triangle inequality states that if a and b are any real numbers, then | a + b | | a | + | b | .

Proof

We prove the following limit law: If lim x a f ( x ) = L and lim x a g ( x ) = M , then lim x a ( f ( x ) + g ( x ) ) = L + M .

Let ε > 0 .

Choose δ 1 > 0 so that if 0 < | x a | < δ 1 , then | f ( x ) L | < ε / 2 .

Choose δ 2 > 0 so that if 0 < | x a | < δ 2 , then | g ( x ) M | < ε / 2 .

Choose δ = min { δ 1 , δ 2 } .

Assume 0 < | x a | < δ .

Thus,

0 < | x a | < δ 1 and 0 < | x a | < δ 2 .

Hence,

| ( f ( x ) + g ( x ) ) ( L + M ) | = | ( f ( x ) L ) + ( g ( x ) M ) | | f ( x ) L | + | g ( x ) M | < ε 2 + ε 2 = ε .

We now explore what it means for a limit not to exist. The limit lim x a f ( x ) does not exist if there is no real number L for which lim x a f ( x ) = L . Thus, for all real numbers L , lim x a f ( x ) L . To understand what this means, we look at each part of the definition of lim x a f ( x ) = L together with its opposite. A translation of the definition is given in [link] .

Translation of the definition of lim x a f ( x ) = L And its opposite
Definition Opposite
1. For every ε > 0 , 1. There exists ε > 0 so that
2. there exists a δ > 0 , so that 2. for every δ > 0 ,
3. if 0 < | x a | < δ , then | f ( x ) L | < ε . 3. There is an x satisfying 0 < | x a | < δ so that | f ( x ) L | ε .

Finally, we may state what it means for a limit not to exist. The limit lim x a f ( x ) does not exist if for every real number L , there exists a real number ε > 0 so that for all δ > 0 , there is an x satisfying 0 < | x a | < δ , so that | f ( x ) L | ε . Let’s apply this in [link] to show that a limit does not exist.

Practice Key Terms 2

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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