By I. Craw

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1 f (x) − f (a) f (x) g(x) − g(a) . 1 as x → ∞. = (Rates of growth) One interest in these results is to see how fast functions grow as x → ∞. This is explored further in the exercises. But important results are: • The function ex increases faster than any power of x. • xα increases faster than any power of log x if α > 0. 6 Taylor’s Theorem We have so far explored the Mean Value theorem, which can be rewritten as f (a + h) = f (a) + hf (c) where c is some point between a and a + h. ] We have already met the approximation f (a + h) ∼ f (a) + hf (a) when we studied the Newton - Raphson method for solving an equation, and have already observed that the Mean Value Theorem provides a more accurate version of this.

Then |f (x) + g(x) − (f (a) + g(a))| ≤ |f (x) − f (a)| + |g(x) − g(a)| < /2 + /2 = . This gives the result. The other results are similar, but rather harder; see (Spivak 1967) for proofs. Note: Just as when dealing with sequences, we need to know that f /g is defined in some neighbourhood of a. This can be shown using a very similar proof to the corresponding result for sequences. 17. Proposition. Let f be continuous at a, and let g be continuous at f (a). Then g ◦ f is continuous at a Proof. Pick > 0.

Exercise. Let f (x) = exp The first point in our study of differentiable functions is that it is more restrictive for a function to be differentiable, than simply to be continuous. 7. Proposition. Let f be differentiable at a. Then f is continuous at a. 2. SIMPLE LIMITS 43 Proof. To establish continuity, we must prove that limx→a f (x) = f (a). 0 x−a as x → a. Hence f is continuous at a. 8. Example. Let f (x) = |x|; then f is continuous everywhere, but not differentiable at 0. Solution. 15 that |x| is continuous.

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