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By Gill G.S.

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0 Part (iii) Consider the inequality Area of triangle ABC ≤ Area of sector ADC ≤ Area of triangle ADE 1 1 sin θ 1 cos θ sin θ ≤ θ ≤ . 2 2 2 cos θ 58 CHAPTER 2. LIMITS AND CONTINUITY Assume that θ is small but positive. Multiply each part of the inequality by 2/ sin θ to get θ 1 cos θ ≤ ≤ . sin θ cos θ On taking limits and using the squeeze theorem, we get lim+ θ = 1. sin θ lim+ sin θ = 1. θ θ→0 By taking reciprocals, we get θ→0 Since sin(−θ) sin θ = , −θ θ sin θ lim− = 1. θ−0 θ Therefore, lim θ→0 sin θ = 1.

Then 1 > 0, 1 < 1 and 1 (1 + |f | + |g(c)|) = 1 = 2(1 + |f (c)| + |g(c)|) < . Since f is continuous at c and 1 > 0, there exists δ1 > 0 such that |f (x) − f (c)| < whenever, |x − c| < δ1 . 1 Also, since g is continuous at c and |g(x) − g(c)| < 1 1 > 0, there exists δ2 > 0 such that whenever, |x − c| < δ2 . Let δ = min{δ1 , δ2 } and |x − c| < δ. For these choices of x, we get |f (x)g(x) − f (c)g(c)| = |(f (x) − f (c) + f (c))(g(x) − g(c) + g(c)) − f (c)g(c)| = |(f (x) − f (c))(g(x) − g(c)) + (f (x) − f (c))g(c) + f (c)(g(x) − g(c))| ≤ |f (x) − f (c)| |g(x) − g(c)| + |f (x) − f (c)| |g(c)| + |f (c)| |g(x) − g(c)| < 1 · 1 + 1 |g(c)| + 1 |f (c)| < 1 (1 + |g(c)| + |f (c)|) , (since 1 < 1) < .

1. 21 Show that (i) lim sin θ = 0 θ→0 sin θ (iii) lim =1 θ→0 θ (ii) lim cos θ = 1 θ→0 1 − cos θ (iv) lim =0 θ→0 θ graph Part (i) By definition, the point C(cos θ, sin θ), where θ is the length of the arc CD, lies on the unit circle. It is clear that the length BC = sin θ is less than θ, the arclength of the arc CD, for small positive θ. Hence, −θ ≤ sin θ ≤ θ and lim sin θ = 0. θ→0+ For small negative θ, we get θ ≤ sin θ ≤ −θ and lim sin θ = 0. θ→0− Therefore, lim sin θ = 0. θ→0 Part (ii) It is clear that the point B approaches D as θ tends to zero.

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