Use limit theorems and/or limit techniques to determine the limits of the following functions. If the limit does not exist, write DNE(Does not exist).

Evaluate the limits of the following functions: 1. lim (5x- + 5x + 15) 1-+3 8. lim x--1 X- 1x-+3x-4 (CE Board Nov. 1997) 2. lim (3x - 5) 9. lim 12-16 (ME Board 1998) 1 +4 X-4 3. lim x-4 X--3x42 10. lim -2 x-2 (ECE Board 1993) 4. lim x-81 sin x X--9 x19 1 1. lim x-0 sin-x (ECE Board April 2002) 5. lim 2-100 12. lim V2+1-4 X-+-10 3410 h 6. lim x3-64 12-169 1-+4 X-4 13. lim 1- 13 *-13 7. lim x+2 15. lim Sinx - cos x 1 +0 x-1 tanx -1Lesson 2.1: These are some of the basic LIMIT LAWS THEOREM 1. Use Direct Substitution to evaluate the limit of ax" + b, where n is a positive integer. lim ax" + b = a(c)" + b Example: lim 3x* - 5 = 3(1)* - 5 = 3 - 5 = - 2 x-1 THEOREM 2. Distribute the limit symbol to evaluate the limit of a sum, difference, product, or quotient. If the lim f(x) and lim g(x) exist, then the following theorems are true. Sum and Difference lim f(x) + g(x) = limf(x) + limg(x) Example 1: lim (3x3 + 2x ) = lim 3x3 + lim 2x = 3(1)3 + 2(1) = 3 + 2 = 5 *-1 x-1 X- 1 lim (3x3 - 2x ) = lim 3x3 - lim 2x = 3(1)3 -2(1) = 3-2 = 1 x-1 Product lim f(x) * g(x) = lim f(x) * limg(x) x-c Example 2: x-1 lim (3x') (2x) = (lim 3x ) ( lim 2x) = [3(1)*][2(1) ] = (3) (2) = 6 Quotient f(x) lim f (x) lim x-+c x-cg(x) lim g(x) . where lim g(x) # 0 Example 3: 3x3 lim 3x lim 3(1)3 = NIW r-+1 2x lim 2x 2(1) , the limit of the denominator # 0. THEOREM 3. Constant Multiple Law 1. lim k * f(x) = klim f(x) 2. lim k = k, where k is any constantlim (x2 + 3x - 4) = 42 + 3(2) -4 = 24 Example 2: Evaluate lim (x + 1)(3x + 1)(6 -9x) from UB Engineering Manual lim (x + 1)(3x + 1)(6 -9x) = (-1+ 1)(3( -1) + 1)(6-9( -1)) = 0 Example 3: Evaluate lim -1 X-+3x+2 x-1 3-1 lim= x-+3x+2 3+2 TECHNIQUE 2: L'Hopital's Rule If DIRECT SUBSTITUTION will result to 0/0 or indeterminate form, we can make use of technique 2. This is by far the most useful technique of evaluating limits of indeterminate forms. Assume that f(x) and g(x) are differentiable on an open interval containing 1 and that f(a) = g(a) =0, also assume that g '(x) #0 (except possibly at a). Then lim J(x) f'(x) lim- x-ag(x) x-ag'(x) Example 1: Evaluate lim, +1 12-x-2 ty Notice that in this example, if we use direct substitution the result is lim X-x-2 _(-1)'-(-1)-2 _1+1-2_0(this is called indeterminate form). r--1 x+1 -1+1 IF THE RESULT IN DIRECT SUBSTITUTION IS 0/0 OR INDETERMINATE, THAT'S THE ONLY TIME WE CAN USE THE L'Hopital's Rule Step 1: Solve for f '(x) and g'(x)[Power Rule, will be discussed later] f(x) = numerator = x- - x - 2 f(x)= x?-x-2 = 1x'-1x-2. Multiply 1 to the exponent 2 then f' (x) = 2x -1 subtract 1 to the exponent 2: 2*1x31 = 2x'. Then remove the variable x in -1x = -1. Disregard the constant-2. Answer ZX - 1 g(x) = denominator = x + 1 g (x ) = 1 g(x) = x + 1= 1x + 1. Since it is only x raised to 1, remove the variable x in 1x that equals to 1. Then disregard the constant + 1. Answer: 1 Step 2: Substitute f '(x) and g ' (x) to the new limit equation. 2x - 1 lim (you can now use direct substitution) = 2(-1) -1 = -3 Answer: lim 1-x-2_ -= - 3 x-+-1 x+1 Example 2: Evaluate lim- 4x2-2x X If we use direct substitution, the result is 0/0 or indeterminate. So we can use technique 2. Step 1: Solve for f '(x) and g '(x) ((x)=4x2 - 2X. Multiply 4 to the exponent 2 f(x) = 4x2 - 2x then subtract 1 to the exponent 2 : 2"4x2-1 = 8x!. f' (x) = 8x - 2 Then remove the variable x in -2x = -2. Answer: 8x - 2 g (x) =x g(x) = x = 1x. Since it is only x raised to 1, 9 (x) =1 remove the variable x in 1x that equals to 1. Answer: 1 Step 2: Substitute f '(x) and g ' (x) to the new limit equation.4x- - 2x Bx - 2 lim = lim = you can now use direct substitution. X X-+0 8x - 2 8(0) - 2 lim - 2 X-+0 YOU MAY USE L'Hopital's Rule TWICE IF IN USING THE FIRST THE RESULT IS STILL INDETERMINATE. TECHNIQUE 3. FACTORING If Direct Substitution will yield to 0/0, you can always use the L'Hopital's Rule. But if you still find that difficult, you may resort to using Factoring. The goal in factoring is to make sure to divide out or remove the factor that will make the equation indeterminate when you attempt to use direct substitution. Let's take a sample. Example 1. Evaluate lim -1 If we attempt to use direct substitution, the result is 0/0 or indeterminate. Notice the denominator x -1 is a factor of the numerator x2 - 1. lim = lim (x - 1)(x + 1) = lim x + 1 ( now you can use direct substitution) mix-1 x - 1 x-+1 lim x + 1 = 1+1 = 2 TECHNIQUE 4. RADICAL FUNCTIONS Most of the time, radical signs confuse students and end up skipping the problem. The key for problems involving radical is consider factoring or multiply by conjugate. You may try any of the two, whichever you see fit. Example 1: Evaluate lim 1-9 VX-3 S Conjugates When the denominator of a rational expression is a binomial that contains a radical, the denominator is Multiply by its conjugate rationalized. This is done by using the of the denominator. The conjugate of a binomial is a binomial having the same two terms with the sign x - 9 v x+3 (x - 9) (vx+ 3) of the second term changed lim- = lim- 19vx - 3 vx+3 x-9 (x -9) The conjugate of 5+ 16 is 5-V6 lim Vx + 3 = V9 + 3 = 6 The conjugate of v3x -/2y is v3x - VZy Source: (Garcia Jessica, Simplifying radical Example 2. Evaluate lim- x-9 expressions, rational exponents, radical X-9 VX-3 equations) Use Factoring lim- x-9 (vx - 3) (vx + 3) = lim x-9 vx - 3 = lim vx + 3 = v9 + 3 = 6 X-9 Vx - 3 TECHNIQUE 5. TRIGONOMETRIC FUNCTIONS. You may attempt to use direct substitution but make sure to use Radian mode if the approaching value of x is very small. If the angle given is in degree, then use degree measure. For limits involving trigonometric functions, you need your notes in Pre-Calculus 1 specifically the different trigonometric identities table. The goal is to factor out and divide the identity that will make the functions indeterminate when using the direct substitution. The L'Hopital's Rule may be used butsolving for the derivative of a trigonometric function is way ahead our schedule in the study of limits. We will go back to using L'Hopital's Rule once we've finished derivative rules. Example 1: Evaluate lim tan x x-0 sin 2x In order to answer this, you may try direct substitution using the radian measure but in this specific problem, it won't work. So we need to use the trigonometric identities in order to divide a common factor. Recall: sin 2x = 2sinxcosx and tanx=sinx/cosx We can now rewrite our equation as lim sinx/cost x-+0 2sinxcosx lim sinx/cosx = lim sinx = lim x-+0 2sinxcosx x-0 2sinxcos-x X-+0 2cos x 2cos-0 Example 2: Evaluate lim sin x x-n 1+cosx "use radian measure since x is approaching IT Recall: sin x = 1- cos2 x sin-x 1 - cos-x lim (1 + cosx) (1 - cosx) lim = lim a n 1 + cosx = lim 1 - cosx = 1 - cost = 2 x a 1 + cosx 1 + cosxExample 4: lim 3(6x2 + 4) = 3 * lim 6x2 + 4 = 3[6(1)2 + 4] = 3(10) = 30 X-1 Example 5: Example 6: lim 3 = 3 lim 8 = 8 x-100 X-6 THEOREM 4. Powers and Roots: if p,q are integers with q#0, then lim If (x) ]" exists and lim If (x) ]" = (lim f(x))" Example 7 lim (2x + 1)= = [lim (2x + 1)]= = [2(1) + 1/2 = 37 = V3 Assuming that lim f(x) 2 0 if q is even, and that lim f(x) # 0 if p/q X-+C 0. In particular, for n a positive integer, lim If (x) ]" = (lim f(x))" lim Vf(x) =" lim f(x) In the second limit, assume that lim f(x) 2 0 if n is even. THEOREM 5. SQUEEZE VALUE THEOREM: Assume that for x# c ( in some open interval containing c). g(x)