Theorems on Limits Example 1: lim(π₯π₯ 2 + 3π₯π₯ + 4) = lim π₯π₯ 2 + lim 3π₯π₯ + lim 4 π₯π₯→2 π₯π₯→2 2 π₯π₯→2 π₯π₯→2 = οΏ½lim π₯π₯οΏ½ + 3 lim π₯π₯ + 4 π₯π₯→2 π₯π₯→2 = [2]2 + 3(2) + 4 = 14 Example 2: lim(π₯π₯ + 4)√2π₯π₯ + 5 = lim(π₯π₯ + 4) lim √2π₯π₯ + 5 π₯π₯→2 π₯π₯→2 π₯π₯→2 = οΏ½lim π₯π₯ + lim 4οΏ½ οΏ½ lim(2π₯π₯ + 5) π₯π₯→2 π₯π₯→2 π₯π₯→2 = οΏ½lim π₯π₯ + lim 4οΏ½ οΏ½ lim 2π₯π₯ + lim 5 π₯π₯→2 π₯π₯→2 π₯π₯→2 π₯π₯→2 = οΏ½lim π₯π₯ + lim 4οΏ½ οΏ½2lim π₯π₯ + lim 5 π₯π₯→2 π₯π₯→2 π₯π₯→2 π₯π₯→2 = (2 + 4)οΏ½2(2) + 5 = 18 Example 3: lim(3π₯π₯ + 4)2 = οΏ½lim(3π₯π₯ + 4)οΏ½ π₯π₯→3 π₯π₯→3 2 = οΏ½lim 3π₯π₯ + lim 4οΏ½ π₯π₯→3 π₯π₯→3 = οΏ½3lim π₯π₯ + lim 4οΏ½ π₯π₯→3 π₯π₯→3 2 2 = [3(3) + 4]2 = 169 NOTE: The limits of the functions in the above examples can be obtained by straight substitution. For instance, in example 2, we see that straight substitution of π₯π₯ = 2 gives the desired limit. Thus in practice, the solution may simply be written as follows: lim(π₯π₯ + 4)√2π₯π₯ + 5 = (2 + 4)οΏ½2(2) + 5 π₯π₯→2 = (6)√9 = 18 Indeterminate Forms Consider the function defined by ππ(π₯π₯) ππ(π₯π₯) = π·π·(π₯π₯) π·π·(π₯π₯) ≠ 0 Suppose at π₯π₯ = ππ, ππ(ππ) = π·π·(ππ) = 0 ππ(ππ) 0 ππ(ππ) = π·π·(ππ) = 0 which is undefined. We say that at π₯π₯ = ππ, the function ππ(π₯π₯) assumes the indeterminate form ∞ 0 0 . The other indeterminate form that we shall encounter here is ∞ . Obtaining any of these forms by straight substitution does not necessarily mean that ππ(π₯π₯) has no limit. We shall see in the examples below that even if ππ(π₯π₯) assumes 0 the indeterminate form 0 at π₯π₯ = ππ, the limit of ππ(π₯π₯) may be definite, i.e., the limit exists. This limit is usually found by changing the expression defined the ππ(π₯π₯) into a form to which the theorems on limits can be used. Consider the following examples. Example 1: Evaluate lim π₯π₯ 2 −4 π₯π₯→2 π₯π₯−2 Solution: This cannot be evaluated by straight substitution since when π₯π₯ = 2, we have π₯π₯ 2 −4 π₯π₯−2 4−4 0 = 2−2 = 0 0 which is meaningless. That is, at π₯π₯ = 2, the function assumes the indeterminate form 0 . However, if π₯π₯ ≠ 2, then π₯π₯ 2 −4 π₯π₯−2 = (π₯π₯−2)(π₯π₯+2) π₯π₯−2 = π₯π₯ + 2 Therefore, to evaluate the limit of the given function, we proceed as follows: lim π₯π₯ 2 −4 π₯π₯→2 π₯π₯−2 = lim π₯π₯→2 (π₯π₯−2)(π₯π₯+2) π₯π₯−2 = lim(π₯π₯ + 2) π₯π₯→2 = 2+2 =4 The example above illustrates the fact that ππ(π₯π₯) may have a limit at a number ππ even though the value ππ(ππ) of the function is undefined. Moreover, it shows that the limit and the value of the function are two different concepts. Example 2: Evaluate lim π₯π₯→2 ππ(π₯π₯)−ππ(2) π₯π₯−2 if ππ(π₯π₯) = π₯π₯ 2 − 3π₯π₯ Solution: 0 A straight substitution of π₯π₯ = 2 leads to the indeterminate form of 0. Since ππ(π₯π₯) = π₯π₯ 2 − 3π₯π₯, then ππ(2) = 22 − 3(2) = 4 − 6 = −2. Hence, lim π₯π₯→2 ππ(π₯π₯)−ππ(2) π₯π₯−2 = lim π₯π₯→2 = lim π₯π₯→2 = lim π₯π₯→2 (π₯π₯ 2 −3π₯π₯)−(−2) π₯π₯−2 π₯π₯ 2 −3π₯π₯+2 π₯π₯−2 (π₯π₯−1)(π₯π₯−2) π₯π₯−2 = lim(π₯π₯ − 1) π₯π₯→2 =1 L’s Hospital’s Rule (LHR) This rule is stated here somewhat casually, without strictly mentioning the specific conditions required. If lim ππ(π₯π₯) = lim ππ(π₯π₯) = 0, then lim π₯π₯→ππ ππ(π₯π₯) π₯π₯→ππ ππ(π₯π₯) π₯π₯→ππ = lim ππ′(π₯π₯) π₯π₯→ππ ππ′(π₯π₯) In words, LHR states that to evaluate the limit of the fraction provided the latter limit exists. ππ(π₯π₯) ππ(π₯π₯) 0 that takes the form 0 at π₯π₯ = ππ, differentiate the numerator and denominator separately and then take the limit of the new fraction 0 ππ′(π₯π₯) ππ′(π₯π₯) . In case, this new fraction assumes again the form 0, the process may be repeated, that is, reapply LHR. However, before reapplying LHR at any stage, we may simplify first the new quotient whenever possible. For instance, simplification by cancellation of common factors of the denominator and numerator may be done first before reapplying LHR. The use of LHR will be illustrated in the next three sections. ππ ∞ Section 1. The Indeterminate Forms ππ and ∞ If lim ππ(π₯π₯) = lim ππ(π₯π₯) = 0, then the function defined by π₯π₯→ππ π₯π₯→ππ ππ(π₯π₯) at π₯π₯ = ππ. To evaluate lim π₯π₯→ππ ππ(π₯π₯) Example 1: Evaluate lim π₯π₯→0 0 ππ(π₯π₯) ππ(π₯π₯) 0 is said to have the indeterminate form 0 = 0 , we apply LHR. Consider the example below. ππ 2π₯π₯ −ππππππππ π₯π₯ Solution: 0 The quotient assumes the form 0 when π₯π₯ = 0. Applying LHR, we get lim π₯π₯→0 ππ 2π₯π₯ −ππππππππ π₯π₯ = lim π₯π₯→0 2ππ 2π₯π₯ +π π π π π π π₯π₯ 1 = 2ππ 0 + π π π π π π 0 = 2(0) + 0 =2 ∞ ∞ If lim ππ(π₯π₯) = lim ππ(π₯π₯) = ∞, then the fraction defined by π₯π₯→∞ π₯π₯→∞ at π₯π₯ = ππ. LHR is also applicable for this form. Example 2: Evaluate lim Solution: ππ(π₯π₯) ππ(π₯π₯) is said to have the indeterminate form π₯π₯ 2 π₯π₯→∞ ππ 2π₯π₯ Since lim π₯π₯ 2 π₯π₯→∞ ππ lim π₯π₯→∞ ∞ = , we apply LHR. Thus 2π₯π₯ π₯π₯ 2 ππ 2π₯π₯ ∞ = lim π₯π₯→∞ = lim 2π₯π₯ 2ππ 2π₯π₯ π₯π₯ π₯π₯→∞ ππ = lim 1 π₯π₯→∞ 2ππ 2π₯π₯ =0 ∞ (= ∞) (cancelling out 2) (reapplying LHR) Section 2. The Indeterminate Forms 0 (±∞) and ∞ − ∞ If lim ππ(π₯π₯) = 0 and lim ππ(π₯π₯) = ±∞, then the fraction defined by the product of ππ(π₯π₯)ππ(π₯π₯) is said to π₯π₯→∞ π₯π₯→∞ 0 ∞ have the indeterminate form 0 (±∞) at π₯π₯ = ππ. Forms of this type can be changed into the form 0 or ∞ so that LHR can be applied. To take effect this change, we rewrite ππ(π₯π₯)ππ(π₯π₯) into any of the following forms: ππ(π₯π₯) 1 ππ(π₯π₯) or ππ(π₯π₯) 1 ππ(π₯π₯) In general, one form is better than the other but the choice will depend upon the given product ππ(π₯π₯)ππ(π₯π₯). Example 1. Evaluate lim π₯π₯π₯π₯π₯π₯π₯π₯π₯π₯ π₯π₯→0 Solution: Let ππ(π₯π₯) = π₯π₯ and ππ(π₯π₯) = ππππππππ. The limit of their product is of the type 0 ∞ since ππ(π₯π₯) = 0 and ππ(π₯π₯) = ∞ at π₯π₯ = 0. Thus lim π₯π₯π₯π₯π₯π₯π₯π₯π₯π₯ = lim π₯π₯→0 π₯π₯ 1 π₯π₯→0 ππππππππ = lim π₯π₯ π₯π₯→0 π‘π‘π‘π‘π‘π‘π‘π‘ = lim 1 π₯π₯→0 π π π π π π 2 π₯π₯ 1 = (1)2 =1 0 (=0) (by LHR) If lim ππ(π₯π₯) = ∞ and lim ππ(π₯π₯) = ∞, then the fraction defined by ππ(π₯π₯) − ππ(π₯π₯) is said to have the π₯π₯→ππ π₯π₯→ππ 0 ∞ indeterminate form ∞ − ∞. This form can be changed into the form 0 or ∞ by algebraic manipulation so that LHR can be applied. Example 2. Evaluate limππ(π π π π π π π π − π‘π‘π‘π‘π‘π‘π‘π‘) π₯π₯→ 2 Solution: ππ ππ This limit is of the type ∞ − ∞ since π π π π π π 2 = ∞ and π‘π‘π‘π‘π‘π‘ 2 = ∞. limππ(π π π π π π π π − π‘π‘π‘π‘π‘π‘π‘π‘) = limππ οΏ½ π₯π₯→ 2 π₯π₯→ 2 ππππππππ − 1−π π π π π π π₯π₯ = limππ οΏ½ π₯π₯→ 2 π₯π₯→ 2 π₯π₯→ 1 2 =0 ππ 2 οΏ½ (Trigonometric Functions) 0 ππππππππ (=0) −π π π π π π π₯π₯ οΏ½ (by LHR) = limππ(ππππππππ) = cot ππππππππ οΏ½ 0−πππππππ₯π₯ = limππ οΏ½ π π π π π π π π Section 3. The Indeterminate Forms ππππ , ππ∞ , ππππππ ∞ππ The function defined by the expression ππ(π₯π₯) ππ(π₯π₯) may, at a certain value of π₯π₯, assume any of the following indeterminate forms: ππππ if ππ(π₯π₯) = 0 and ππ(π₯π₯) = 0 ππ∞ if ππ(π₯π₯) = 1 and ππ(π₯π₯) = ∞ ∞ππ if ππ(π₯π₯) = ∞ and ππ(π₯π₯) = 0 To evaluate lim ππ(π₯π₯) ππ(π₯π₯) when any of the indeterminate forms above is obtained, we may perform the π₯π₯→ππ following steps; 1. Let ππ = lim ππ(π₯π₯) ππ(π₯π₯) π₯π₯→ππ 2. Take the natural logarithm of both sides of the equation in (1). ππππ ππ = ππππ lim ππ(π₯π₯) ππ(π₯π₯) π₯π₯→ππ = lim ππππ ππ(π₯π₯) ππ(π₯π₯) (Laws of Logarithms) π₯π₯→ππ (Laws of Logarithms) = lim ππ(π₯π₯) ππππ ππ(π₯π₯) π₯π₯→ππ Thus The limit at this point takes the form 0 (±∞) and we, therefore, apply the method used in Section 2. 3. ππππ ππ = lim π₯π₯→ππ ππππ ππ(π₯π₯) 1 ππ(π₯π₯) 0 ∞ . This limit is of the type 0 or ∞ . 4. Apply LHR to the right member of (3). 5. Suppose ππππ ππ = πΏπΏ. Then ππ = ππ πΏπΏ where πΏπΏ is any real number. Therefore lim ππ(π₯π₯)ππ(π₯π₯) = ππ πΏπΏ π₯π₯→ππ Example. 2 π₯π₯ Evaluate lim οΏ½1 + οΏ½ π₯π₯ π₯π₯→∞ Solution: This limit is of the type 1∞ . 2 π₯π₯ Let ππ = lim οΏ½1 + π₯π₯οΏ½ π₯π₯→∞ 2 π₯π₯ Then ππππ ππ = lim ππππ οΏ½1 + π₯π₯οΏ½ π₯π₯→∞ 2 = lim π₯π₯ ππππ οΏ½1 + π₯π₯οΏ½ π₯π₯→∞ = lim π₯π₯→∞ = lim π₯π₯→∞ οΏ½ 1 π₯π₯→∞ 2 1 π₯π₯ −2 2οΏ½οΏ½ π₯π₯2 οΏ½ 1+ π₯π₯ −1 π₯π₯2 = lim οΏ½ = 2 π₯π₯ ππππ οΏ½1+ οΏ½ 2 1+ οΏ½ 2 π₯π₯ 1+0 =2 Hence ππ = ππ 2 and we conclude that 2 π₯π₯ lim οΏ½1 + οΏ½ = ππ2 π₯π₯ π₯π₯→∞ (= ∞ β 0) 0 (= ) 0 (by LHR) Gillesania’s Principle for Limits That is; To evaluate the lim ππ(π₯π₯), substitute for π₯π₯ a value that is very close to ππ then use a calculator. π₯π₯→ππ For π₯π₯ For π₯π₯ For π₯π₯ For π₯π₯ → 2, substitute π₯π₯ = 1.9999 or π₯π₯ = 2.0001 → −5, substitute π₯π₯ = −4.9999 or π₯π₯ = −5.0001 → 0, substitute π₯π₯ = 0.00001 → ∞, substitute π₯π₯ = 99999 Consider the following examples: Example 1: π₯π₯π₯π₯π₯π₯ ππππ −ππππ ππ→ππ ππ−ππ Substitute π₯π₯ = 2.9999 (2.9999)4 −81 (2.9999)−3 ππ Example 2: π₯π₯π₯π₯π₯π₯ οΏ½ − ππ ππ→ππ ππ = 107.99 ≈ 108 ππππ −ππ οΏ½ Substitute π₯π₯ = 0.0001 1 1 οΏ½0.0001 − ππ 0.0001−1οΏ½ = 0.49999 ≈ 0.5 Example 3: π₯π₯π₯π₯π₯π₯ οΏ½ ππ→ππ ππππππππ ππππππππππ οΏ½ Substitute π₯π₯ = 0.0001° ππππππ(0.0001°) οΏ½ππππππ[2(0.0001°)]οΏ½ = 2 ππππ +ππ−ππ −ππππ −ππ Example 4: π₯π₯π₯π₯π₯π₯ οΏ½ ππ→ππ ππππππππ ππ−ππππ οΏ½ Substitute π₯π₯ = 0.0001 ππππππππππππ Note: set your calculator to radian mode ππ 0.0001 +ππ −0.0001 −(0.0001)2 −2 π π π π π π 2 (0.0001)−(0.0001)2 = −0.25
