Limits Definition 1 (Limit) . If the values f ( x ) of a function f : - PDF document

Limits Definition 1 (Limit) . If the values f ( x ) of a function f : A → B get very close to a specific, unique number L when x is very close to, but not necessarily equal to , a limit point c , we say the limit of f ( x ), as x approaches c , is L and write lim x → c f ( x ) = L . The number c is referred to as the limit point . Note: This is not a rigorous definition. It is really a preliminary attempt at a definition and will be replaced later by a more rigorous definition. One obvious problem is that the concept of closeness is subjective. Some limits are relatively easy to calculate. Example: lim x → 5 ( x 2 + 3). It’s obvious that x 2 will be close to 25 and x 2 + 3 will be close to 28 when x is close to 5, so lim x → 5 ( x 2 + 3) = 28. 3 t + 9 Example: lim t → 7 t − 5 . It’s obvious that 3 t is close to 21, 3 t + 9 is close to 30, t − 5 is close to 2 and thus 3 t + 9 t − 5 is close to 30 3 t + 9 2 = 15 when t is close to 7, so lim t → 7 t − 5 = 15. Most of the limits that come up naturally are not quite so obvious and easy to calculate, but often yield to a little algebraic manipulation and common sense. x 2 + x − 12 . If one examines the numerator and denominator of x 2 + x − 12 Example: lim x → 3 x − 3 x − 3 for values of x close to 3, it is obvious both are close to 0 and one cannot immediately estimate the quotient. Such limits are referred to as indeterminate . This particular type of indeterminate is sometimes referred to symbolically as the 0 0 case. We say symbolically because, of course, there is no quotient 0 0. It is a short way of referring to the fact that we are trying to find the limit of a quotient where both the numerator and the denominator are getting very close to 0. However, one may simplify x 2 + x − 12 , finding x 2 + x − 12 = ( x − 3)( x + 4) = x + 4 x − 3 x − 3 x − 3 when x � = 3. Since x + 4 is clearly close to 7 when x is close to 3, so must x 2 + x − 12 and x − 3 x 2 + x − 12 it follows that lim x → 3 = 7. x − 3 We have implicitly made use of the following theorem, which is an almost immediate consequence of the preliminary definition of a limit. Theorem 1. If f ( x ) = g ( x ) for x � = c and either f or g has a limit at c , then both must have a limit at c and their limits must be the same. In effect, this is a reiteration of the fact that the value of a function at a limit point has no effect on a limit. We use this theorem in the calculation of the limit almost every time we are faced with an indeterminate. The calculation generally looks like the following: lim x → c f ( x ) = · · · = lim x → c g ( x ) = L . 1

2 The idea is that we start with an indeterminate f ( x ) and keep simplifying, each time getting either a new form of f ( x ) or another function that is equal to f ( x ) except at c , until we have found some function g ( x ) whose limit is easy to find. For example, the calculation in the previous example might look like the following: x 2 + x − 12 ( x − 3)( x + 4) lim x → 3 = lim x → 3 = lim x → 3 ( x + 4) = 7. x − 3 x − 3 To repeat, in a nutshell, we usually wind up calculating limits of indeterminates by sim- plifying until we’ve found a function, equal to the original except at the limit point, which is not indeterminate . Most of the time, the algebraic manipulation we need to do to simplify is in one of the following three categories. (1) Factor and cancel. (2) Rationalize and cancel. (3) Simplifying a complex fractional expression The example we saw fell in the first category, factor and cancel . Consider the following

3 1 z − 1 1 z − 1 z − 4 · 4 z 4 4 Technique 2: lim z → 4 z − 4 = lim z → 4 4 z 4 − z − 1 4 z = − 1 = lim z → 4 4 z ( z − 4) = lim z → 4 16. Here, we observe the complications arose from the denominators of 4 and z in the numer- ator and simply multiplied by 1 in the form 4 z 4 z . There will be other manipulations that will come up, and sometimes the simplication involved using these techniques may not be quite as straightforward, but most of the examples we run across will fall into one of these three cases. Different Types of Limits Besides ordinary, two-sided limits, there are one-sided limits (left-hand limits and right- hand limits), infinite limits and limits at infinity. One-Sided Limits √ x 2 − 4 x − 5. lim x → 5 One might think that since x 2 − 4 x − 5 → 0 as x → 5, it would follow that √ x 2 − 4 x − 5 = 0. lim x → 5 √ But since x 2 − 4 x − 5 = ( x − 5)( x + 1) < 0 when x is close to 5 but x < 5, x 2 − 4 x − 5 is undefined for some values of x very close to 5 and the limit as x → 5 doesn’t exist. But √ x 2 − 4 x − 5 is close to 0 when x is close to 5 and x > 5, we would still like a way of saying so we say the Right-Hand Limit exists, write lim x → 5 + √ x 2 − 4 x − 5 = 0 √ x 2 − 4 x − 5 approaches 0 as x approaches 5 from the right. and say Sometimes we have a Left-Hand Limit but not a Right-Hand Limit. Sometimes we have both Left-Hand and Right-Hand Limits and they’re not the same. Sometimes we have both Left-Hand and Right-Hand Limits and they’re equal, in which case the ordinary limit exists and is the same.  x 2 if x < 1   x 3 f ( x ) = if 1 < x < 2  x 2 if x > 2 .  lim x → 1 − f ( x ) = lim x → 1 + f ( x ) = 1, so the left and right hand limits are equal and lim x → 1 f ( x )1. lim x → 2 − f ( x ) = 8 while lim x → 2 + f ( x ) = 4, so the left and right hand limits are different and lim x → 2 f ( x ) doesn’t exist. Limits at Infinity 2 x Suppose we’re interested in estimating about how big x + 1 is when x is very big. It’s 2 x 2 x 2 2 x easy to see that x + 1 = x ) = if x � = − 1 and thus x + 1 will be very close to 2 x (1 + 1 1 + 1 x if x is very big. We write 2 x lim x →∞ x + 1 = 2 2 x and say the limit of x + 1 is 2 as x approaches ∞ .

4 2 x Similarly, x + 1 will be very close to 2 if x is very small and we write 2 x lim x →−∞ x + 1 = 2 2 x and say the limit of x + 1 is 2 as x approaches −∞ . Here, small does not mean close to 0, but it means that x is a negative number with a large magnitude ( absolute value ). A convenient way to find a limit of a quotient at infinity (or minus infinity) is to factor out the largest term in the numerator and the largest term in the denominator and cancel what one can. 5 x 2 − 3 lim x →∞ 8 x 2 − 2 x + 1 = x 2 (5 − 3 x 2 ) lim x →∞ x 2 ) = x 2 (8 − 2 x + 1 5 − 3 = 5 x 2 lim x →∞ 8 − 2 x + 1 8 x 2 5 x − 3 lim x →∞ 8 x 2 − 2 x + 1 = x (5 − 3 x ) lim x →∞ x 2 ) = x 2 (8 − 2 x + 1 5 − 3 x lim x →∞ x 2 ) = 0 x (8 − 2 x + 1 Infinite Limits 1 If x is close to 1, it’s obvious that ( x − 1) 2 is very big. We write 1 lim x → 1 ( x − 1) 2 = ∞ 1 and say the limit of ( x − 1) 2 is ∞ as x approaches 1 . 1 Similarly, lim x → 1 − ( x − 1) 2 = −∞ . Technically, a function with an infinite limit doesn’t actually have a limit. Saying a function has an infinite limit is a way of saying it doesn’t have a limit in a very specific way. Infinite limits are inferred fairly intuitively. If one has a quotient f ( x ) g ( x ), one may look at how big f ( x ) and g ( x ) are. For example, If f ( x ) is close to some positive number and g ( x ) is close to 0 and positive, then the limit will be ∞ . If f ( x ) is close to some positive number and g ( x ) is close to 0 and negative, then the limit will be −∞ . If f ( x ) is close to some negative number and g ( x ) is close to 0 and positive, then the limit will be −∞ . If f ( x ) is close to some negative number and g ( x ) is close to 0 and negative, then the limit will be ∞ .