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INTRODUCTION In the previous lesson, we presented and illustrated the limit theorems. We started recalling these limit theorems. Theorem 1. Let c, k, L and Al be real numbers, and let f(x) and g(x) be functions defined on some open interval containing c, except possibly at c. 1. If lim f(r) exists, then it is unique. That is, if lim f(r) = L and lim f(r) = M, then I +C I++C L = M. 2. lim c = c. I-+C 3. lim r = c4. Suppose lim S(z) = L and lim g(r) = M. i. (Constant Multiple) lim[k . g(x)] = & . M. ii. (Addition) lim[S(r) +g(r)] = Lt M. iii. (Multiplication) lim|f ()9()) - LA. iv. (Division) lim I (r) AT. Provided M $ 0. v. (Power) lim[f(r)]" = D' for p. a positive integer. vi. (noot/ Radical) lim Vf(x) - VL for positive integers n, and provided that L > 0 when n is even. In this lesson, we will show how these limit theorems are used in evaluating algebraic functions. Particularly, we will illustrate how to use them to evaluate the limits of polynomial, rational and radical functions. Topic: Limit of Polynomial Function We start with evaluating the limit of polynomial function. EXAMPLE 1: Determine lim (2x + 1). Solution. From the theorems above, lim (2r + 1) = lim 2x + lim 1 (Addition) c-+1 I++1 = (2 lim x ) + 1 (Constant Multiple) = 2(1) + 1 lim z = c) =2+1 = 3. EXAMPLE 2: Determine lim (2r - 4x? +1). Solution. From the theorems above, lim (2r3 - 4r' + 1) = lim 2x- lim 4r + lim 1 (Addition) 14-1 I4-1 = 2 lim x - 4 lim x' + 1 14-1 (Constant Multiple) 14-1 = 2(-1)3 - 4(-1)2+1 (Power) =-2-4+1 =-5. EXAMPLE 3: Evaluate lim (3r* - 2x - 1). 1-+0 Solution. From the theorems above, lim (3r* - 2x - 1) = lim 3r - lim 2r - lim 1 (Addition) I+40' 1-+0 = 3 lim r - 2 lim x2 - 1 (Constant Multiple) 1-+0 1-+0 = 3(0) - 2(0) - 1 (Power) =0-0-1 = -1.Topic: Limit of Rational Function We will now apply the limit theorems in evaluating rational functions. In evaluating the limits of such functions. recall from Theorem 1 the Division Rule, and all the rules stated in Theorem 1 which have been useful in evaluating limits of polynomial functions, such as the Addition and Product Rules. EXAMPLE 4: Evaluate lim= Solution. First. note that lim r = 1. Since the limit of the denominator is nonzero, we can apply the Division Rule. Thus. lizn 1 1 - (Division) lim a EXAMPLE 5: Evaluate lim - 1-21 + 1 Solution. We start by checking the limit of the polynomial function in the denominator. lim (x + 1) = lim r + lim 1 = 2+1 = 3. I-+2 Since the limit of the denominator is not zero, it follows that lim I I lim r-21+1 lim (z + 1) (Division) EXAMPLE 6: Evaluate lim (I - 3)(=2-2) First, note that I2 +1 lim (x2 | 1) - lim r | lim 1 - 1 | 1- 20. Thus, using the theorem, (2 3) (=2 2) lim (x - 3)(x2 - 2) lim 12 + 1 lim (z2 + 1) (Division) lim ( - 3) . lim (12 - 2) (Multipication) lim z - lim 3 lim x2 - lim 2 = 2 (Addition) (1 - 3) (12 - 2) 2 = 1. 1-51 EXAMPLE 7 Evaluate lim 241 1+ 312 + 414 Solution. Since the denominator is not zero when evaluated at I = 1, we may apply Theorem 3: 1- 51 1-5(1) lim -4 141 1+ 3x2 +41 1+ 3(1)2 + 4(1) 0. Therefore, by the Radical/Root Rule, lim VI = lim = = VI = 1. EXAMPLE 9 Evaluate lim va + 4. 1-+0 Solution. Note that lim (z + 4) = 4 > 0. Hence, by the Radical/Root Rule, lim vr + 4 = lim (r + 4) = V4 = 2. 1-+0 EXAMPLE 10 Evaluate lim Vx2 + 3r - 6. I -+ - 2 Solution. Since the index of the radical sign is odd, we do not have to worry that the limit of the radicand is negative. Therefore, the Radical/Root Rule implies that or EXAMPLE "1 Evaluate lim V21 + 5 1 +2 1 - 31 Solution. First, note that lim (1 - 3r) = -5 0. Moreover, lim (2z + 5) = 9 > 0. Thus, using the Division and Radical Rules of Theorem 1, we obtain V2:T. 1 5 lim v2r + 5 lim (2x + 5) lim I-+2 I-2' z-+2 1 - 31 lim 1 - 3r -5A. Written Work/s: ACTIVITY: Evaluate the following limits. Show your solutions. 1. lim (3.r - 5) lim (2x - 4x* + x - 2) 2. 3. lim lim " - 1 4. (21 + 1) (3 +3) 5. lim lim (3.r - 5) 6. lim (274 - 403 + 2 -2) 7 8. lim r-2 1 1 9. lim 2r 10. lim (2.r + 1) (12 + 3) r- -3