Lecture 5: Finding limits analytically Simple indeterminate forms

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1 Lecture 5: Finding its analytically Simple indeterminate forms Objectives: (5.) Use algebraic techniques to resolve 0/0 indeterminate forms. (5.) Use the squeeze theorem to evaluate its. (5.3) Use trigonometric techniques to resolve 0/0 indeterminate forms. Functions that differ at a point The substitution technique that we saw in the last lecture is the simplest and most useful of all the it techniques we will see. However, it has two major drawbacks. The first drawback is theoretical: the substitution technique often creates very serious confusion regarding the it concept. It is etremely important to remember that in general, a it is not about what happens at a point, but rather what happens near the point. The behavior of the function at the it point is irrelevant when determining a it. Having said that, it is ironic that the substitution technique actually requires us to evaluate the function at the it point. You should not think about the substitution technique as function evaluation, but rather as the end result of our it laws. The second drawback is practical: the substitution technique can, and often does, fail, even when a it eists. Consider these eamples: 4 = 4 sin = Substitution is useless because neither epression is defined at the corresponding it point. Nonetheless the its eist (as we shall seen very soon). In both cases, direct substitution produces the form 0/0. 0/0 is not a number and it does not simplify to a number, but it is what each of the epressions looks like when we substitute the it point for the variable. In one case, the 0/0 form is associated with the it 4, and in the other case, the 0/0 form is associated with the it. Because 0/0 forms can be resolved to give different its (or none at all), we say that 0/0 is an indeterminate form. When evaluating a it, if substitution results in the indeterminate form 0/0, then no conclusion can be made without doing more work. The it may or may not eist. We will see several ways to resolve indeterminate forms. Most of our strategies will rely on the following theorem. Theorem Functions that differ at a point Suppose f() = g() for all c on an open interval containing c. If g() eists, then f() c c also eists and f() = g(). c c To understand the value of this theorem, think about how we simplified epressions in our earlier algebra courses. For eample, 4 = ( )( + ) ( ) = //////////( ) + ) ( ) ////////// = +,. Since the epression on the far left is not defined when =, equality certainly does not hold at =, but it holds everywhere else. The functions on the left and right differ only at the single point =. According to the theorem, 4 = ( + ) = 4. The most common applications of Theorem make use of the following algebraic strategies: 4

2 . Factor and cancel (as above). Epand and simplify 3. Multiply by the conjugate and simplify 4. Get a common denominator and add (or subtract) The following eamples illustrate these strategies in order. Eample Evaluate: more work. Let s try factoring. The functions = ( + )( 3) ( + )( + 3) = ////////// ( + ) ( 3) = ////////// ( + ) ( + 3) = 5 f() = 6 and g() = are identical everywhere ecept at =. By Theorem, they have the same its everywhere, in particular at =. Eample Evaluate: h 0 ( + h) h more work. Let s use the fact that ( + h) = + h + h. h 0 ( + h) h Is it clear how we are using Theorem? h + h h/( + h) = = = ( + h) = h 0 h h 0 h/ h 0 Eample 3 Evaluate: r + r 3 more work. Rationalizing the numerator will help. Since we only epect cancellation in the numerator, we will only carry out the multiplication there. r + r 3 r + + r + + = r + 4 ( r + + )(r 3) = Is it clear how we are using Theorem? (r 3) ///////// ( r + + )(r 3) ///////// = 4 Eample 4 Evaluate: Once again direct substitution yields the indeterminate form 0/0. We cannot draw a conclusion without more work. In this case, it will help to get a common denominator and carry out the subtraction in the numerator = 4 ( + 4) 4( + 4) = 4( + 4) = / 4( + 4)/ = 6 5

3 The squeeze theorem Theorem is a comparison theorem two functions are compared and information about one is obtained from the other. The net theorem is also a comparison theorem. It doesn t have the practical importance of Theorem, but it can be quite useful. Theorem Squeeze theorem Suppose f, g, and h are functions for which g() f() h() for all in an open interval containing c, ecept possibly at c. If g() = L and h() = L, c c then f must have the same it at c: f() = L. c In the theorem, the values of f are squeezed between the values of g and h. If g and h are tending to a common it, then f is going with them. Eample 5 Suppose f is a function with the property that for all. What can be said about f()? + 4 f() Let g() = + 4 and h() = The graphs of g and h are shown below. Notice that g() = h() = (by direct substitution). 8 y 6 y=h() y=g() Since f is squeezed between g and h, it follows that f() =. The squeeze theorem can be used to prove the following useful result. Theorem 3 A special it sin = 6

4 Justification for Theorem 3 is given on the net page. This result is more general than it seems at first glance. You should interpret it as saying: For eample, sin g() g() sin 5 5 = The net eamples illustrate Theorem 3. Eample 6 as g() 0. sin( ) cos Use Theorem 3 to evaluate. sin( 4) = = 4 more work. Let s multiply by the conjugate of the numerator. cos Now use it laws to rewrite... + cos + cos = sin cos ( + cos) = sin + cos = 0 = 0. sin ( + cos). Eample 7 Evaluate: sin 7 Direct substitution yields 0/0. More work! Since sin 7 7 sin, we cannot simplify sin 7. However, it looks like Theorem 3 should be useful. We can use it after multiplying by 7/7. sin7 = 7 7 sin 7 = 7 sin 7 7 = 7 = 7 Eample 8 Evaluate: tan 3 8 Direct substitution yields 0/0. More work! tan 3 = sin 3 cos3 = 3 8 sin3 3 cos3 = 3 8 = 3 8 7

5 Some justification for Theorem 3... h() Area of small triangle Area of sector Area of big triangle sin ( + h())sin or sin + h() Now as 0, it should be pretty clear that h() 0. Therefore we get It follows that 0 sin sin = By the it laws, the it of the reciprocal function is the reciprocal of the it: sin = 0 8