By I. Craw

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We use the above criterion. Note that f (1) = 1. Also lim f (x) = lim x2 = 1 while x→1− x→1− lim f (x) = lim x = 1 = f (1). x→1+ x→1+ so f is continuous at 1. 14. Exercise. Let f (x) = Show that f is continuous at 0. [In fact x cos x for x ≥ 0. f is continuous everywhere]. 15. Example. Let f (x) = |x|. Then f is continuous in R . Solution. Note that if x < 0 then |x| = −x and so is continuous, while if x > 0, then |x| = x and so also is continuous. It remains to examine the function at 0. From these identifications, we see that limx→0− |x| = 0+, while limx→0+ |x| = 0+.

In the next example, it looks as though it is going to be hard, but turns out to be quite possible. 8. Example. Let f (x) = Then f is continuous at 0. x  0 if x = 0. CHAPTER 4. LIMITS AND CONTINUITY 34 Solution. We prove this directy from the definition, using that fact that, for all x, we have | sin(x) ≤ 1|. Pick > 0 and choose δ = [We know the answer, but δ = /2, or any value of δ with 0 < δ ≤ will do]. Then if |x| < δ, x sin 1 1 1 − 0 = x sin = |x|. sin ≤ |x| < δ ≤ x x x as required. Note that this is an example where the product of a continuous and a discontinuous function is continuous.

Proposition. Let g(x) = 1/f (x). Then g(x) → 0+ as x → ∞ iff f (x) → ∞ as x → ∞. Let y = 1/x. Then y → 0+ as x → ∞; conversely, y → ∞ as x → 0+ Proof. Pick > 0. We show there is some K such that if x > K, then 0 < y < ; indeed, simply take K = 1/ . The converse is equally trivial. 6 Continuity on a Closed Interval So far our results have followed because f is continuous at a particular point. In fact we get the best results from assuming rather more. Indeed the results in this section are precisely why we are interested in discussing continuity in the first place.

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Advanced Calculus And Analysis by I. Craw


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