Limit laws 2

Two-sided and one-sided limits

  • From the definition of two-sided and one sided limits, we get the following statement.
    • Theorem. Let \(f(x)\) be a function defined around but possibly not at a number \(a\). Then we have \(\lim_{x\to a}f(x)=L\) if and only if \(\lim_{x\to a^-}f(x)=L=\lim_{x\to a^+}\).
    • That is, the two-sided limit exists if and only if both one-sided limits exist and they agree.
  • For example, we can use this to compute \(\lim_{x\to0}|x|\).
    • Since if \(x>0\) then \(|x|=x\), we have \(\lim_{x\to0^+}|x|=\lim_{x\to0^+}(x)=0\).
    • Similarly, since if \(x<0\) then \(|x|=-x\), we have \(\lim_{x\to0^-}|x|=\lim_{x\to0^-}(-x)=0\)
    • The two together show that \(\lim_{x\to0}|x|=0\).
  • Similarly we can consider \(\lim_{x\to0}\frac{|x|}{x}\).
    • We have \(\lim_{x\to0^+}\frac{|x|}{x}=\lim_{x\to0^+}\frac{x}{x}=\lim_{x\to0^+}1=1\).
    • Also, we have \(\lim_{x\to0^-}\frac{|x|}{x}=\lim_{x\to0^-}\frac{-x}{x}=\lim_{x\to0^-}(-1)=-1\).
    • Since although the two one-sided limits exist, they have different values, the two-sided limit \(\lim_{x\to0}\frac{|x|}{x}\) does not exist.

More examples

  • The greatest integer function is defined as follows.
    • The value \([[x]]\) is the largest integer that is \(\le x\). For example, we have \([[2]]=2\), \([[2.5]]=2\), \([[e]]=2\), \([[3]]=3\), \([[-2.5]]=-3\).
    • Consider the limit \(\lim_{x\to3}[[x]]\).
    • If \(3<x<4\), then we have \([[x]]=3\). Therefore, we have \(\lim_{x\to3^+}[[x]]=\lim_{x\to3^+}3=3\).
    • If \(2<x<3\), then we have \([[x]]=2\). Therefore, we have \(\lim_{x\to3^-}[[x]]=\lim_{x\to3^-}2=2\).
    • Since the two one-sided limits don't agree, the two-sided limit \(\lim_{x\to3}[[x]]\) does not exist.
    • Exercises. 2.3: 42,46,54.

Limits and comparison

  • Knowledge about how the values of two or three functions compare shows us how their limits compare.
    • Theorem. Suppose that \(f(x)\le g(x)\) when \(x\) is near \(a\), and that the limits of both functions at \(a\) exist. Then we have \(\lim_{x\to a}f(x)\le\lim_{x\to a}g(x)\).
    • Squeeze theorem. Suppose that \(f(x)\le g(x)\le h(x)\) when \(x\) is near \(a\), and that \(\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L\). Then we have \(\lim_{x\to a}g(x)=L\).
    • The Squeeze theorem is used a lot when you want to compute a limit involving trigonometric functions.
    • This is because the range of \(\cos(x)\) and \(\sin(x)\) is \([-1,1]\), which is equivalent to \(-1\le\cos(x),\sin(x)\le1\).
    • For example, let us consider \(\lim_{x\to0}(x^2\sin\frac{1}{x})\).
    • Note that you can't use the Product law, since \(\lim_{x\to0}\sin\frac{1}{x}\) does not exist, as we've seen on Monday.
    • Instead, we can apply the Squeeze theorem: from \(-1\le\sin\frac{1}{x}\le1\) we get \(-x^2\le x^2\sin\frac{1}{x}\le x^2\), and we have \(\lim_{x\to0}(-x^2)=\lim_{x\to0}(x^2)=0\).
    • Therefore, we get \(\lim_{x\to0}x^2\sin\frac{1}{x}=0\).
    • Exercise. 2.3: 40.


Continuity at a number

  • Definition. Let \(f(x)\) be a function and let \(a\) be a number. Then \(f(x)\) is continuous at \(a\), if we have \(\lim_{x\to a}f(x)=a\).
    • Note that this requires three things.
      1. The value \(f(a)\) is defined, that is the number \(a\) is in the range of the function \(f(x)\).
      1. The limit \(\lim_{x\to a}f(x)\) exists.
      1. We have \(\lim_{x\to a}f(x)=f(a)\).
    • If any of these statements does not hold, that is \(f(x)\) is not continuous at \(a\), then we say that \(f(x)\) is discontinuous at \(a\), or that \(f(x)\) has a discontinuity at \(a\).
    • Exercises. 2.5: 10a-d, 12, 18, 24
    • We say that \(f(x)\) has a jump discontinuity at \(a\), if the one-sided limits at \(a\) exist, but we have \(\lim_{x\to a^-}f(x)\ne\lim_{x\to a^+}f(x)\). An example of this is 2.5.10d: you get the jumps either from that you can't pay money in smaller quantities than cents or that the taxi calculator measures the distance numerically, that is up to a certain precision.
    • We say that \(f(x)\) has an infinite discontinuity at \(a\), if we have \(\lim_{x\to a^-}f(x)=\pm\infty\) or \(\lim_{x\to a^+}f(x)=\pm\infty\). An example of this is 2.5.18.
    • We say that \(f(x)\) has a removable discontinuity at \(a\), if the limit \(\lim_{x\to a}f(x)\) exists, but either \(f(a)\) is not defined, or we have \(\lim_{x\to a}f(x)\ne f(a)\). You can remove the discontinuity by setting the value \(f(a)\) to \(\lim_{x\to a}f(x)\). An example of this is 2.5.24.
    • Click here for plots. (Note that I graphed \(f(x)=[[x]]\) instead of real-life taxi prices.)