Limit of Product¶

We find a rule for calculating the limit of the product of two functions, as in $$ \lim_{x \to a} \Bigl( f(x)g(x) \Bigr). $$ This is similar to the limit of sum, but more difficult.

"Error functions"¶

(The title is in quotes, because the term "error function" often refers to a different thing.)

Suppose that $$ \lim_{x \to a} f(x) = 3. $$ Let's define the following function: $$ E_f(x) = f(x)-3 $$ This "error function" measures how far away the function is from its limit, and in which direction. For example, if $f(x) = 2.99$, then $E_f(x) = -0.01$, which means that $f(x)$ is quite close to its limit, and less than its limit.

Let's now take the limit of $E_f$ using the limit of sum: $$ \begin{align} \lim_{x \to a} E_f(x) &= \lim_{x \to a} (f(x)-3) \\ &= \lim_{x \to a} (f(x) + (-3)) \\ &= \lim_{x \to a} f(x) + \lim_{x\to a} (-3) \end{align} $$ The limit of $f(x)$ is $3$, and according to limit of a constant, the limit of $-3$ is simply $-3$, so we get: $$ \lim_{x \to a} E_f(x) = 3 + (-3) = 0. $$ This shouldn't be surprising: "limit is zero" is basically a precise way to say "small".

It is also possible to rearrange the definition $E_f(x) = f(x)-3$ into this: $$ f(x) = 3 + E_f(x) $$ This should also make sense. It says that $f(x)$ consists of the limit value $3$ and a small additional error.

We will use these "error functions" below. Of course, the limit doesn't have to be $3$: this trick can be used whenever a function has any limit.

Simple case: both limits are positive integers¶

Suppose that $$ \lim_{x \to a} f(x) = 3 \quad \text{and} \quad \lim_{x \to a} g(x) = 5. $$ We calculate the limit of $f(x)g(x)$.

Let's start by defining the following "error functions": $$ \begin{align} E_f(x) &= f(x)-3 \\ E_g(x) &= g(x)-5 \end{align} $$ And rearranging them: $$ \begin{align} f(x) &= 3 + E_f(x) \\ g(x) &= 5 + E_g(x) \end{align} $$ We can now start calculating the limit of $f(x)g(x)$: $$ \begin{align} \lim_{x \to a} \Bigl( f(x)g(x) \Bigr) &= \lim_{x \to a} (3+E_f(x))(5+E_g(x)) \\ &= \lim_{x \to a} \Bigl(15 + 3E_g(x) + 5E_f(x) + E_f(x)E_g(x) \Bigl) \end{align} $$ We ended up with 4 things added, so we can use the limit of sum to split them into separate limits: $$ \begin{align} \lim_{x \to a} \Bigl( f(x)g(x) \Bigr) &= \lim_{x \to a} 15 + \lim_{x \to a} 3E_g(x) + \lim_{x \to a} 5E_f(x) + \lim_{x \to a} \bigl(E_f(x)E_g(x)\bigr) \end{align} $$ The first three limits are easy:

• The limit of $15$ is simply $15$ by limit of a constant.
• The limit of $3E_g(x)$ can be calculated as a limit of sum: $$ \begin{align} \lim_{x \to a} 3E_g(x) &= \lim_{x \to a} \Bigl(E_g(x)+E_g(x)+E_g(x)\Bigr) \\ &= \lim_{x \to a} E_g(x)+\lim_{x \to a} E_g(x)+\lim_{x \to a} E_g(x) \\ &= 0+0+0 \\ &= 0 \end{align} $$
• The limit of $5E_g(x)$ can be calculated similarly, and it is also zero.

We now have: $$ \lim_{x \to a} \Bigl( f(x)g(x) \Bigr) = 15 + \lim_{x \to a} \bigl( E_f(x)E_g(x) \bigr) $$ The limit of $E_f(x)$ is zero, so if $x$ is close enough to $a$, we have $E_f(x) \approx 0$ with tolerance $0.1$. (This works with any tolerance. Let's just pick $0.1$). This means that $$ -0.1 < E_f(x) < 0.1. $$ Multiplying by $E_f(x)$ brings numbers closer to zero. For example, if $E_f(x) = -0.05$, then $E_f(x) \cdot 40 = -2$, and $-2$ is much closer to zero than $40$. (Also, the sign flips if $E_f(x)$ is negative.)

Let's now look at $E_f(x)E_g(x)$. The limit of $E_g(x)$ is zero, which means that for any tolerance $t > 0$, we have $E_g(x) \approx 0$ when $x$ is close enough to $a$. However, multiplying by $E_f(x)$ brings the value even closer to zero, so we also have $E_f(x)E_g(x) \approx 0$ with tolerance $t$. Therefore $$ \lim_{x \to a} \bigl( E_f(x)E_g(x)\bigr) = 0. $$

This result shouldn't feel very surprising. It says that if $f(x)$ is close to 3 and $g(x)$ is close to 5, then $f(x)g(x)$ is close to 15.

Of course, the same reasoning works with any positive integers instead of 3 and 5. We need positive integers for the step where we did $3E_g(x) = E_g(x)+E_g(x)+E_g(x)$.

They don't need to be integers¶

Suppose that $$ \lim_{x \to a} f(x) = 3.14 \quad \text{and} \quad \lim_{x \to a} g(x) = 6.9. $$ We calculate the limit of $f(x)g(x)$.

The above calculation mostly works as is, but this part needs to be modified a bit: $$ \begin{align} \lim_{x \to a} 3E_g(x) &= \lim_{x \to a} \Bigl(E_g(x)+E_g(x)+E_g(x)\Bigr) \\ &= \lim_{x \to a} E_g(x)+\lim_{x \to a} E_g(x)+\lim_{x \to a} E_g(x) \\ &= 0+0+0 \\ &= 0 \end{align} $$ The problem is that we get $3.14E_g(x)$ instead of $3E_g(x)$. We can still split it up similarly, but $0.14 E_g(x)$ will remain. $$ \begin{align} \lim_{x \to a} 3.14E_g(x) &= \lim_{x \to a} \Bigl(E_g(x)+E_g(x)+E_g(x) + 0.14E_g(x)\Bigr) \end{align} $$ However, multiplying by $0.14$ makes numbers smaller because it's less than $1$, so the limit of $0.14E_g(x)$ is also zero: $$ \begin{align} \lim_{x \to a} 3.14E_g(x) &= \lim_{x \to a} \Bigl(E_g(x)+E_g(x)+E_g(x) + 0.14E_g(x)\Bigr) \\ &= 0+0+0+0 \\ &= 0 \end{align} $$ We will also need the limit of $6.9E_f(x)$, and that's zero for the same reason, so we get $$ \lim_{x \to a} \Bigl( f(x)g(x) \Bigr) = 3.14 \cdot 6.9 = 21.666. $$ This works with any positive numbers instead of $3.14$ and $6.9$.

They don't need to be positive¶

Our limit calculation works fine if one or both of the limits is zero. In that case, we get something like $\lim_{x \to a} 0E_g(x)$, which is actually the limit of a constant.

For a negative limit, we get something like $$ \lim_{x \to a} -3.14E_g(x). $$ We already know that the limit of $3.14E_g(x)$ will be zero (see above). The minus sign in front doesn't matter: if $3.14E_g(x) \approx 0$ with some tolerance, then also $-3.14E_g(x) \approx 0$ with the same tolerance, because the minus sign flips to the opposite side of zero without affecting the distance. Therefore $$ \lim_{x \to a} -3.14E_g(x) = 0 $$ and the calculation works as expected.

This is known as the rule for the limit of a product.

Example¶

With the limit rules we have so far, we can calculate e.g. the following limit: $$ \lim_{x \to 2} (2x^2+4x+7) $$ Let's start with limit of sum: $$ \lim_{x \to 3} (2x^2+4x+7) = \lim_{x \to 3} 2x^2 + \lim_{x \to 3} 4x + \lim_{x \to 3} 7 $$ Consider the limit of $2x^2$. Let's view $2x^2$ as the product of 3 things, as in $2 \cdot x \cdot x$. We use the rule for limit of product. Just like the rule for sum, it can be applied to more than two functions, like this: $$ \begin{align} \lim_{x \to 3} 2x^2 &= \lim_{x \to 3} (2 \cdot x \cdot x) \\ &= \left( \lim_{x \to 3} 2 \right)\left( \lim_{x \to 3} x \right)\left( \lim_{x \to 3} x \right) \end{align} $$ We can handle the limit of $4x$ similarly. Finally, we can use $\lim_{x \to 3} x = 3$ (derived here) and limit of a constant. Like this: $$ \begin{align} \lim_{x \to 3} (2x^2+4x+7) &= \lim_{x \to 3} 2x^2 + \lim_{x \to 3} 4x + \lim_{x \to 3} 7 \\ &= \left(\lim_{x \to 3} 2\right)\left( \lim_{x \to 3} x \right)^2 + \left(\lim_{x \to 3} 4 \right)\left( \lim_{x \to 3} x \right) + \lim_{x \to 3} 7 \\ &= 2 \cdot 3^2 + 4 \cdot 3 + 7 \\ &= 37. \end{align} $$ We can check this by plugging in a number near 3: $$ 2 \cdot (3.001)^2 + 4 \cdot 3.001 + 7 = 37.016002 $$