Limit of division¶

We find a rule for calculating the limit of two functions divided by each other, as in $$ \lim_{x \to a} \frac{f(x)}{g(x)}. $$ We start with a simple case and then generalize the result.

Simple case¶

Suppose that $\lim_{x \to a} g(x) = 4$. We calculate the limit of $\frac{1}{g(x)}$.

This doesn't divide by zero if $x$ is close enough to $a$. To see why, let's take a small enough tolerance, e.g. $0.1$. If $x$ is close enough to $a$, we now get $g(x) \approx 4$ with tolerance $0.1$, which means that $3.9 < g(x) < 4.1$, so $g(x)$ is surely positive.

We suspect that the limit of $\frac{1}{g(x)}$ is probably $\frac{1}{4}$, because $g(x)$ is close to $4$. Let's compare $\frac{1}{g(x)}$ and $\frac{1}{4}$ by subtracting them. (We used a similar trick to derive the limit of product.) $$ \begin{align} \frac{1}{g(x)} - \frac{1}{4} &= \frac{4}{4g(x)} - \frac{g(x)}{4g(x)} \\ &= \frac{4-g(x)}{4g(x)} \end{align} $$ Let $t > 0$ be any tolerance. On top, we have $4-g(x)$. If $x$ is close enough to $a$, we have $g(x) \approx 4$ with tolerance $t$, which means that the distance between $g(x)$ and $4$ is less than $t$: $$ -t < 4-g(x) < t $$ On bottom, we have $4g(x)$. If $x$ is close enough to $a$, we have $g(x) > 3.9$, so $$ 4g(x) > 4 \cdot 3.9 = 15.6. $$ This means that $4g(x)$ is so big that dividing by it brings numbers closer to zero, and hence anything that is between $-t$ and $t$ will stay between $-t$ and $t$: $$ -t < \frac{4-g(x)}{4g(x)} < t $$ Because $\frac{4-g(x)}{4g(x)} = \frac{1}{g(x)} - \frac{1}{4}$, we get: $$ -t < \frac{1}{g(x)} - \frac{1}{4} < t $$ This shows that $\frac{1}{g(x)} \approx \frac{1}{4}$ with tolerance $t$ when $x$ is close enough to $a$, so $$ \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{4}. $$

Slightly more general case¶

Unfortunately, the above doesn't work very nicely if we replace $4$ with other numbers. This is the part that causes problems: $$ 4g(x) > 4 \cdot 3.9 = 15.6 $$ Instead of $15.6$, we might get a number that is smaller than 1. Then dividing by it doesn't bring things closer to zero as expected. For example, dividing by $0.5$ is same as multiplying by two. A negative limit value would also cause problems.

These problems could be fixed or worked around (feel free to try if you want), but IMO it's not worth the effort, because there is an easier way. Suppose that the limit of $g(x)$ is any nonzero number $B \ne 0$. The idea is to adjust the function $g(x)$ so that instead of $B$, we get $4$.

Let's try dividing the original function by its limit $B$. Using limit of product, we get $$ \begin{align} \lim_{x \to a} \frac{g(x)}{B} &= \lim_{x \to a} \frac{1}{B} g(x) \\ &= \left( \lim_{x \to a} \frac{1}{B} \right) \left( \lim_{x \to a} g(x) \right) \\ &= \frac{1}{B} \cdot B = 1. \end{align} $$ If we also multiply by 4, we get $$ \begin{align} \lim_{x \to a} \frac{4g(x)}{B} &= \lim_{x \to a} \frac{4}{B} g(x) \\ &= \left( \lim_{x \to a} \frac{4}{B} \right) \left( \lim_{x \to a} g(x) \right) \\ &= \frac{4}{B} \cdot B = 4. \end{align} $$ Therefore, if we can get the constant $\frac{4}{B}$ in front of $g(x)$, we know how to calculate the limit: $$ \lim_{x \to a} \frac{1}{\frac{4}{B}g(x)} = \frac{1}{4} $$ It is quite easy to get the constant in front, because in division, you can multiply the top and bottom by anything nonzero you want, and it doesn't affect anything. Like this: $$ \begin{align} \lim_{x \to a} \frac{1}{g(x)} &= \lim_{x \to a} \frac{\frac{4}{B}}{\frac{4}{B}g(x)} \\ &= \lim_{x \to a} \frac{4}{B} \frac{1}{\frac{4}{B}g(x)} \\ &= \left( \lim_{x \to a} \frac{4}{B} \right) \left( \lim_{x \to a} \frac{1}{\frac{4}{B}g(x)} \right) \\ &= \frac{4}{B} \cdot \frac{1}{4} \\ &= \frac{1}{B}. \end{align} $$ Here we used limit of product and limit of constant.

General case¶

Any division $\frac{f(x)}{g(x)}$ can be viewed as a multiplication $f(x) \frac{1}{g(x)}$. In this case, that's useful because we can apply limit of product: $$ \begin{align} \lim_{x \to a} \frac{f(x)}{g(x)} &= \lim_{x \to a} \left( f(x) \frac{1}{g(x)} \right) \\ &= \left( \lim_{x \to a} f(x) \right) \left( \lim_{x \to a} \frac{1}{g(x)} \right) \\ &= \left( \lim_{x \to a} f(x) \right) \frac{1}{\displaystyle\lim_{x \to a} g(x)} \\ &= \frac{\displaystyle\lim_{x \to a} f(x)}{\displaystyle\lim_{x \to a} g(x)} \end{align} $$