By Gill G.S.

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**Example text**

Part (v) Suppose that M > 0 and lim g(x) = M . Then we show that x→c lim x→c 1 1 = . 1. INTUITIVE TREATMENT AND DEFINITIONS 47 Since M/2 > 0, there exists some δ1 > 0 such that M 2 M 3M − + M < g(x) < 2 2 M 3M 0< < g(x) < 2 2 1 2 < |g(x)| M |g(x) − M | < whenever 0 < |x − c| < δ1 , whenever 0 < |x − c| < δ1 , whenever 0 < |x − c| < δ1 , whenever 0 < |x − c| < δ1 . Let > 0 be given. Let 1 = M 2 /2. Then δ > 0 such that δ < δ1 and > 0 and there exists some 1 |g(x) − M | < 1 whenever 0 < |x − c| < δ < δ1 , M − g(x) |g(x) − M | 1 1 = = − g(x) M g(x)M |g(x)|M 1 1 = · |g(x) − M | M |g(x)| 1 2 < · · 1 M M 21 = 2 M = whenever 0 < |x − c| < δ.

For this reason, many cycles of the sine wave pass from the value −1 to the value +1 and a rapid oscillation occurs near zero. None of the following limits exist: lim+ sin x→0 1 x , lim− sin x→0 1 x , lim sin x→0 1 x . It is not possible to define the function f at 0 to make it continuous. This kind of discontinuity is called an “oscillation” type of discontinuity. 7 Consider f (x) = x sin 1 x as x tends to 0. graph In this example, sin 1 , oscillates as in Example 6, but the amplitude x 40 CHAPTER 2.

Part (i) First, we show that for all real c, lim sin θ = sin c or equivalently lim | sin θ − sin c| = 0. θ→c θ→c We observe that θ+c θ−c sin 2 2 (θ − c) ≤ 2 sin 2 0 ≤ | sin θ − sin c| = 2 cos sin (θ−c) 2 = |(θ − c)| (θ−c) 2 Therefore, by squeeze theorem, 0 ≤ lim | sin θ − sin c| ≤ 0 · 1 = 0. θ−c It follows that for all real c, sin θ is continuous at c. Next, we show that lim cos x = cos c or equivalently lim | cos x − cos c| = 0. x→c x→c We observe that 0 ≤ | cos x − cos c| = −2 sin ≤ |θ − c| sin x−c 2 x−c 2 ; x+c (x − c) sin 2 2 sin x+c ≤1 2 Therefore, 0 ≤ lim | cos x − cos c| ≤ 0 · 1 = 0 x→c and cos x is continuous at c.