Limit Properties¶

If we know that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist, then do we have $$ \lim_{x \to a} (f(x) + g(x)) = \left( \lim_{x \to a} f(x) \right) + \left( \lim_{x \to a} g(x) \right)? $$ A rule like this would be handy for calculating limits. With this, we could for example calculate $$ \lim_{x \to 2} (x+5) = \left( \lim_{x \to 2} x \right) + \left( \lim_{x \to 2} 5 \right) = 2+5=7, $$ where we used most of the results from the limit basics page. On this page, we figure out whether this and other similar things work.

Interval thing¶

Before talking more about limits, we need a result about intervals. Suppose that $a$ and $b$ are numbers in the open interval $(-r, r)$, where $r$ is a positive number. What can we say about $a+b$?

Because $a$ and $b$ are in $(-r, r)$, we have $-r < a < r$ and $-r < b < r$. By combining these inequations together, we get $$ (-r)+(-r) < a+b < r+r, $$ which simplifies to $$ -2r < a+b < 2r. $$ This can be written as $$ a+b \in (-2r, 2r). $$ So, if we have $a,b \in (-r, r)$, then we also know that $a+b \in (-2r, 2r)$. We'll need this result next.

Usually a more general theorem called triangle inequality is used instead of this, but I'm using this instead because it feels somewhat more intuitive to me.

Limit of sum¶

Let's go back to the question about limits. Suppose that $\lim_{x \to a} f(x) = A$ and $\lim_{x \to a} g(x) = B$. We prove that $\lim_{x \to a} (f(x)+g(x)) = A+B$.

Let $I_y$ be an open interval centered around $A+B$. We can write $I_y$ as $(A+B-\epsi, A+B+\epsi)$ for some positive number $\epsi$. By definition of limit, there are input intervals centered around $a$ such that we can fit the values of $f$ and $g$ into any output intervals around $A$ and $B$ respectively. We choose the output intervals $\left( A-\frac{\epsi}{2}, A+\frac{\epsi}{2} \right)$ and $\left( B-\frac{\epsi}{2}, B+\frac{\epsi}{2} \right)$. The corresponding input intervals are drawn below.

In the past we have used the fact that intervals like this overlap, but we can say more than that. One of the intervals is smaller and the other is bigger, and the smaller interval is a subset of the bigger interval; this means that the bigger interval contains all numbers of the smaller interval.

Let $I_x$ be the smaller input interval (if the input intervals are the same size, then they are the same interval, and it doesn't matter which of them is $I_x$). Now any number in $I_x$ is in both of the input intervals, so we have $$ \begin{align} &f(\text{any number in $I_x$ except $a$}) \in \left( A-\frac{\epsi}{2}, A+\frac{\epsi}{2} \right), \\ &g(\text{any number in $I_x$ except $a$}) \in \left( B-\frac{\epsi}{2}, B+\frac{\epsi}{2} \right). \end{align} $$ Letting $x$ be any number in $I_x$ except $a$, we can write this as $$ f(x) = A+p \quad \text{and} \quad g(x)=B+q $$ for some numbers $p,q \in (-\frac{\epsi}{2},\frac{\epsi}{2})$. Now we get $$ f(x)+g(x) = A+B+\underbrace{p+q}_{\in (-\epsi,\epsi)}, $$ where for the underbraced part, we used the above interval thing with $r=\frac{\epsi}{2}$ and $2r=\epsi$. We have $$ f(x)+g(x) \in (A+B-\epsi, A+B+\epsi) = I_y. $$

Limit of product¶

Suppose that $\lim_{x \to a} f(x) = A$ and $\lim_{x \to a} g(x) = B$. I'll show you most of the proof for $\lim_{x \to a} f(x)g(x) = AB$.

Let $I_y=(AB-\epsi,AB+\epsi)$ be an open interval centered around $AB$. By doing same things as for the sum above, we get an input interval $I_x$ such that for any $x$ in $I_x$ except $a$, we get $$ f(x)=A+p \quad \text{and} \quad g(x)=B+q, $$ where we can fit $p$ and $q$ into any interval $(-r,r)$; above we had $p,q \in (-\frac{\epsi}{2}, \frac{\epsi}{2})$ but we could have used any other positive number $r$ instead of $\frac{\epsi}{2}$. Just like in the sum proof, it is now enough to show that by choosing $r$ wisely, we get $$ f(x)g(x) = AB+\text{something in $(-\epsi,\epsi)$}. $$ By expanding $f(x)g(x)$, we get $$ \begin{align} f(x)g(x) &= (A+p)(B+q) \\ &= A(B+q)+p(B+q) \\ &= AB+\underbrace{Aq+pB+pq}_{\text{Want: $\in (-\epsi,\epsi)$}}. \end{align} $$ Fitting $Aq+pB+pq$ into $(-\epsi,\epsi)$ can be done by choosing the $r$, but filling in all the details results in a mess every time I do it, so I'll skip most of it here. However, I'll show you the ideas needed for doing that, and you might be able to fill in the skipped details after reading this if you are interested in them.

To fit all of $Aq+pB+pq$ into $(-\epsi,\epsi)$, it is enough to ensure that $Aq$, $pB$ and $pq$ are all in $(-\frac{\epsi}{3}, \frac{\epsi}{3})$. This can be seen by generalizing the interval thing in the beginning of this page for three numbers instead of two.

Let's say that we want to make the $Aq$ part small enough, as in $Aq \in (-\frac{\epsi}{3}, \frac{\epsi}{3})$. If $A$ is positive, then we can pick $r = \frac{\epsi}{3A}$, because using the fact $q \in (-r,r)$ gives $-r < q < r$, which is same as $$ -\frac{\epsi}{3A} < q < \frac{\epsi}{3A}. $$ Multiplying both sides by $A$ (assuming that it's positive) gives $$ -\frac{\epsi}{3} < Aq < \frac{\epsi}{3}, $$ so $Aq \in (-\frac{\epsi}{3}, \frac{\epsi}{3})$.

Some of the mess comes from handling zero and negative values of $A$ and $B$, and the rest comes from combining the conditions ensuring that each of $Aq$, $pB$ and $pq$ are small enough.

Limit of difference¶

Suppose that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist. By noticing that $f(x)-g(x) = f(x) + (-1)g(x)$, we get $$ \begin{align} \lim_{x \to a} (f(x)-g(x)) &= \lim_{x \to a} \Bigl( f(x) + (-1)g(x) \Bigr) \\ &= \lim_{x \to a} f(x) + \lim_{x \to a} \Bigl( (-1)g(x) \Bigr) \\ &= \lim_{x \to a} f(x) + \underbrace{\Bigl( \lim_{x \to a} (-1) \Bigr)}_{-1}\Bigl( \lim_{x \to a} g(x) \Bigr) \\ &= \lim_{x \to a} f(x) - \lim_{x \to a} g(x). \end{align} $$ Here we used most of the results that we have proved so far:

• Limit of sum: we can take the limits of $f(x)$ and $(-1)g(x)$ separately.
• Limit of product: we can take the limits of $-1$ and $g(x)$ separately.
• Limit of just a number (from the limit basics page): the limit of $-1$ is $-1$.

Limit of division¶

Suppose that $\lim_{x \to a} f(x) = A$ and $\lim_{x \to a} g(x) = B$. Is it true that $$ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{A}{B}? $$ This is not true if $B=0$, because then we would divide by zero. We'll come back to the case $B \ne 0$ later.