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

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.)