Q. 82 - Calculus | StudyQuestionHub

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Q. 82

Question

Write a delta–epsilon proof that shows that the function $f (x) = 2 x + 1$ is continuous at $x = 5$ . (This exercise depends on Section 1.3.)

Step-by-Step Solution

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Answer

Hence we proved $\lim_{x \to 5} (2 x + 1) = 11$ .

1Step 1. Given Information

We are given the function $f (x) = 2 x + 1$ is continuous at $x = 5$ and we need to write a delta–epsilon proof.

2Step 2. Delta–epsilon proof

$f (x) = 2 x + 1 c = 5 L = 2 (5) + 1 = 10 + 1 = 11$

For $f (x) = 2 x + 1$ ,

If $0 < |x - 5| < δ$ we have,

$|(2 x + 1) - 11| = |2 x - 10| = |2 (x - 5)| < 2 δ = 2 (\frac{ε}{2}) = ε ∴ \lim_{x \to 5} (2 x + 1) = 11$

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