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3. LIMIT OF FUNCTION 47 3. Order: Suppose limx→a f (x) = l and limx→a g(x) = k. If f (x) ≥ g(x) for x close to a, then l ≥ k. Conversely, if l > k, then f (x) > g(x) for x close to a. 4. Sandwich: Suppose f (x) ≤ g(x) ≤ h(x) and limx→a f (x) = limx→a h(x) = l. Then limx→a g(x) = l. 5. Composition: Suppose limx→a g(x) = b, limy→b f (y) = c. If g(x) = b for x near a, or f (b) = c, then limy→a f (g(x)) = c. When something happens near a, we mean that there is δ > 0, such that it happens when 0 < |x − a| < δ.

For a function f (x), limx→a f (x) = l if and only if limn→∞ f (xn ) = l for any sequence {xn } satisfying xn = a and limn→∞ xn = a. Proof. We prove the case a and l are finite. The proof for the other cases is similar. Suppose limx→a f (x) = l. Suppose {xn } is a sequence satisfying xn = a and limn→∞ xn = a. Then for any > 0, there are δ > 0 and N , such that 0 < |x − a| < δ =⇒ |f (x) − l| < . n > N =⇒ |xn − a| < δ. The assumption xn = a further implies |xn − a| > 0. 3. LIMIT OF FUNCTION 57 and we conclude limn→∞ xn = a.

X . . . . . .. . . . ........................................................................................................................................................ O ... A B .. 18), we get limx→0+ sin x = 0. By the composition rule, limx→0− sin x = lim−x→0− sin(−x) = − limx→0+ sin x = 0. 3, lim sin x = 0. 3. LIMIT OF FUNCTION 51 This further implies lim cos x = lim 1 − 2 sin2 x→0 x→0 x x = 1 − 2 lim sin x→0 2 2 2 = 1.