University:
University of California, Berkeley
Course:
MATH 56 Linear Algebra
Academic year:

2022
Views:

161

Pages:

5
Author:

SpaceCadet101

One-sided Limits When we consider lim f (x), x→x0 we need to consider the values of the function f on an interval surrounding both sides of x0 . This may be impossible if x0 is an endpoint of the domain of f , or may at the very least be problematic if f is defined piecewise, with different definitions on each side of x0 (making it difficult to algebraically manipulate f ). In order to overcome these difficulties, we can define one-sided limits, where we only consider x values on one side of x0 . When we write x → 2 it means x approaches 2. Similarly, x % 2 means x increases to 2 and x & 2 means x decreases to 2. On a number line, if x % 2 (is increasing to 2), then x approaches 2 from the left side, and if x & 2 (is decreasing to 2), then x approaches 2 from the right side. Definition 0.1.1 (Right-Hand Limit). We say that L is the right-hand limit of f (x) at x0 , written lim f (x) = L x&x0 if for every number ε > 0, there exists a corresponding number δ > 0 such that for all x with x0 < x < x0 + δ we have |f (x) − L| < ε. Definition 0.1.2 (Left-Hand Limit). We say that L is the left-hand limit of f (x) at x0 , written lim f (x) = L x%x0 if for every number ε > 0, there exists a corresponding number δ > 0 such that for all x with x0 − δ < x < x0 we have |f (x) − L| < ε. The above definitions are identical to ε-δ definition of the limit except that now we only require that the δ-interval extend on one side of x0 . Because these definitions are otherwise identical to the ε-δ definition, it follows that all of the rules we previously established for combining two-sided limits (and the sandwich theorem) still hold for one-sided limits. There is one additional result for relating one-sided and two-sided limits. Theorem 0.1.1 (One-sided and Two-sided Limits). A function f has a limit L at x0 if and only if both its left-hand and right-hand limits at x0 are L. The if and only if in this theorem states an equivalence - having a two-sided limit is the same thing as having equal one-sided limits. Although this may seem like a very simple fact, it is indeed useful for showing that limits do not exist. Rather than trying to show all possible limit values do not work, we simply need to show that the one-sided limits are different. If the one-sided limits exist but disagree, then it is impossible for the function to approach a single value as x → x0 , which implies that the two-sided limit does not exist. |x| does not exist. x→0 x Solution Recall that |x| −1 x < 0 = 1 x>0 x Example 1. Prove that lim 1 In order to prove the limit does not exist, we simply need to show that the one-sided limits as x % 0 and x & 0 either disagree, or that one does not exist. Looking from just the left or right side of the point x0 = 0, we have two constant functions. Since the limit of a constant is just that constant, it follows that |x| |x| lim = 1 and lim = −1. x&0 x x%0 x It follows that limx→0 |x|/x does not exist. Note that we have not just shown that the function does not have a specific limit value as x → 0; we have shown that it does not have any limit value. Example 2. Evaluate lim f (x) and lim f (x) for x→0 x→2 f (x) =  2  x 2cos(1/x) 3x −5x−2 x−2   14(√x2 +12−4) x−2 0 sin(x) > cos(x). x Since limx&0 cos(x) = 1, it follows using the sandwich theorem (for one-sided limits) that lim x&0 sin(x) = 1, x and because the left-hand limit equals the right-hand limit, we have that lim sin(x) = 1. x→0 4 cos(x) − 1 = 0. x Solution Since the limit of the denominator is zero we cannot use the quotient rule. Given that we’ve just proven (through a reasonable amount of work) a similar limit, we might try to rewrite cosine in terms of sine, using an appropriate trigonometric identity. In order to remove the 1 from the numerator, we use the half-angle identity Example 4. Show that lim x→0 cos(x) = 1 − 2 sin2 (x/2). We find that −2 sin2 (x/2) sin2 (x/2) cos(x) − 1 = lim = − lim . x→0 x→0 x→0 x x (x/2) lim From here we are getting much closer to something of sin(x)/x. The only problem is we have sin(x/2). However, we also have (x/2) in the denominator. If we let h = (x/2), then we are looking at sin2 (h) sin(h) lim = lim · sin(h). x→0 2h→0 (h) (h) Here we have written everything in terms of h, and have gotten very close to a familiar limit. We just need to notice that as 2h → 0, we have h → 0, so we find sin(h) · sin(h) = 1 · 0 = 0. 2h→0 (h) lim Thus, cos(x) − 1 = 0. x→0 x lim 5