MATH 116, LECTURES 10 & 11: Limits

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1 MATH 6, LECTURES 0 & : Limits Limits In application, we often deal with quantities which are close to other quantities but which cannot be defined eactly. Consider the problem of how a car s speedometer determines (and displays) your speed. If you look at the speedometer and find you are going 52 km/h, you interpret that to mean that, at that instant, you are travelling 52 km/h (i.e. that if you remained at that speed for one hour, you would travel 52 km). Even as we say this, however, we notice a problem in interpretation. What does it mean to travel at a speed at an instant? By definition, an instant is not a length of time. Over any given instant, we travel zero km! Of course, this is not what we really mean when we talk about our speed at a particular instance. Without even thinking about it, what we mean is the best or closest approimation to the instantaneous value by using short intervals around the instance. In other words, we really mean an average speed which accurately represents how fast we feel we are going at the time. In order for the average speed to feel accurate, we should take it over a short time interval. As the interval gets shorter and shorter, we epect the average velocity to get closer and closer to the instantaneous velocity. The observation that things can get closer and closer to meaningful values without ever attain them is formalized mathematically in the concept of a it. Definition.. Consider a function f(). We define the it of and a to be L = a f(). This notation means that, as a, we have f() L. Note: Usually, if we can evaluate at the it value = a, that is the value we are looking for and we are done. (There are counter-eamples for this, however.) The method of resolving its where we cannot evaluate the it value = a will vary depending on the it but the technique remains the same. In general, what we want to do is manipulate the it until we can

2 get into a point where we can evaluate at = a and get a well-defined value. Eample : Evaluate Solution: Naively, we attempt to plug in the it value = 2 to get (2) 2 (2) 2 (2) 2 = 0 0 =? This is called an indeterminate form. This means eactly what it sounds like it should be mean, that we cannot determine the value based on this information. In fact, there are zero over zero its which take the value negative infinity, positive infinity, and every finite value in between! So what do we do now? Well, let s (also naively) consider taking values on either side of = 2 which get increasingly close to the it point (which we know we cannot evaluate). We have y y It certain appears that, as gets closer and closer to two, f() gets closer and closer to three (also see Figure ). Is there some way we can show this is actually the case? In this particular case, the iting problem can be resolved by factoring the numerator. We have 2 2 ( + )( 2) = = ( + ) = Eample 2: Evaluate Solution: Again, we naively evaluate at = 0 to get (0) (0) 2 = 0 0 =? 2 2

3 f()= Figure : The function f() = 2 2 behaves like + everywhere 2 ecept for at = 2 where there is a hole punched in it. We need to find a way to resolve the it, but we cannot factor this epression. Instead we rationalize the root by multiplying both the numerator and denominator by the conjugate of the numerator (the numerator with the sign switched by subtraction to addition). We have ( ) = One-Sided Limits = 0 2 ( ) 2 = 0 2 ( ) = = = 6. For many functions, the iting value of f() is different depending on which way we approach the iting value of. This gives rise to one-sided its. 3

4 Definition 2.. The one-sided it of f() from the left is defined as L = a f(). The one-sided it of f() from the right is defined as L + = a + f(). Eample : Evaluate the left-sided and right-sided its of the following function at = : f() =. Solution: We know that need to consider two cases:. For slightly less than one, > 0 and < 0. becomes unbounded near =. We 2. For slightly greater than one, > 0 and > 0. It follows that 0 =, 0 + =. An -value for which f() is called a vertical asymptote. Eample 2: Evaluate the left-sided and right-sided its of the heaviside function at = 0. Solution: We have U() = { 0, for < 0, for 0. By definition, if we take any values to the left of zero, the heaviside function returns zero, and if we take any values to the right of zero, the heaviside function returns one. It follows that U() = 0, 0 U() =

5 Figure 2: The it of at = is when approaching from the left and when approaching from the right. 3 Squeeze Theorem Suppose we know that L() f() U(). That is to say, we know that the function f() is bounded from below by a lower function L() and above by an upper function U(). Suppose as that L() = U() = L. a a That is to say we know that the upper and lower functions have the same it at = a. What can we say about the it as a for f()? The answer should be obvious (consider Figure 3). We must have f() = L. a There is simply nowhere else for f() to go! This is called the Squeeze theorem. Eample: Use the fact that, for (0, π/2) sin() tan() to prove that sin() 0. 5

6 L =a U() f() L() Figure 3: If L() f() U() on an interval, and a U() = a L() = L for an a in that interval, then we have a f() = L. Solution: We want to find bounds on the epression sin()/ near = 0. In this case, this is a little tricky. We are fortunately given the bounds sin() tan() (which we have not derived and instead take for granted). From the first inequality, this gives us sin() and from the second inequality was have sin() cos() = cos() sin(). We can now clearly see that cos sin() sin() = cos() sin() =. 0 + It follows by the Squeeze theorem that sin() =. 0 + The inequality for the it as 0 follows similarly by substituting (try it!). Otherwise, we are done. 6

7 4 Limits at infinity We are also interested in what happens to functions as the dependent variable approaches infinity. Consider, for instance, a population growth function given by N(t) = 000(0)t + (0) t where N denotes the population size. A key qualitative feature of this interpreting this model is the long-term behaviour, that is to say, that behaviour as time goes on and on. Does the the population eplode, die out, or settle to a finite value (and if so, which value)? Mathematically, we epress this as t N(t). For this eample, we can quickly see that 000(0) t 000(0) t t + (0) t = t + (0) t ( ) (0) t (0) t = t 000 (0) t + = 000. So the population gets closer and closer to 000 as time goes on! Many of the same techniques for evaluating its as finite values as evaluating its at infinity. The most generally technique, however, will be to take terms which approach infinity as the independent variable grows and make it disappear instead. This will be illustrated with a few eamples. Eample : Evaluate Solution: We can see that as, we have /. This is another one of out indeterminant forms. We need to find a way to resolve this indeterminancy. We will do so by dividing the top and bottom by the largest power on the top or bottom, which is 2. We have ) = ( = =

8 Notice that the it could be evaluated because we turned each term which went off to infinity in the original it into something which went to zero. Notice also that, even though, the function approached a finite value. Such a it is called a horizontal asymptote. Eample 2: Evaluate Solution: Notice that, if we naively evaluate as, we have. This is yet another one of our indeterminant forms. We need to resolve this indeterminancy. In this case, we multiply top and bottom (even though there is no bottom to begin with!) by the conjugate of We have = ( ) = ( 2 = = = = 2 ( ) ) Formal Definition of a Limit 2 = =. There is some imprecision in the definition of a it as we have been using it so far. What does it mean for a value to get closer and closer to another value? How close do we have to get before we are satisfied that our answer is the one we want? If there is one thing that mathematicians hate, it is being imprecise (it keeps us up at night!) so mathematicians of years past have given its a more precise definition. 8

9 Definition 5. (Formal definition of a it). Consider a function f(). We will say that a f() = L if, for every ɛ > 0, there eists a δ > 0 such that a < δ implies f() L < ɛ. () This is called either the formal definition of a it or the ɛ-δ definition. These kinds of ɛ-δ definitions of concepts are common in higher year mathematics courses (especially analysis courses). Love them or hate them, they are here to stay! (Although they rarely factor in applied courses, it is still important to understand the intuition behind them.) There are few things worth noting about this definition:. Equation () says that as you get closer in the -direction, you get closer in the y-direction. This corresponds with our intuition of what its are. 2. Not only do we get closer, the fact that we can choose any ɛ > 0 means we can choose ɛ to be arbitrarily small. This means we can get arbitrarily close in the direction. 3. This definition makes concrete the notion of closeness! If we want to be within ɛ of the it value L, we have to pick an value within δ of a. Eample : Prove that using the formal definition of a it = Solution: We need to show that, for any ɛ > 0, there is a δ > 0 such that ( ) < δ implies (3 + 4) < ɛ. Consider (3 + 4) < ɛ. We have that (3 + 4) = = 3 +. In order to make this less than ɛ, i.e. 3 + < ɛ, we need to have + < ɛ/3. This is the δ we need to pick! That is to say, if we pick δ = ɛ/3, we have satisfied the requirements. To see why this is the case, let s start from the beginning. Take ɛ > 0 to be arbitrary (for the interesting case, take it to be arbitrarily small). Pick δ = ɛ/3. 9

10 Now suppose that + < δ = ɛ/3. It follows that (3 + 4) < 3 + < 3(ɛ/3) = ɛ. This was the implication we needed in order for the it to be satisfied! This means that = holds by the rigorous standards set by mathematicians of years gone by (although it was probably just obvious to the rest of us). 0