Limit of Sum¶

If we know that $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist, then do we have $$ \lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)? $$ A rule like this would be useful for calculating limits. With this, we could for example calculate $$ \lim_{x \to 2} (x+5) = \lim_{x \to 2} x + \lim_{x \to 2} 5 = 2+5=7, $$ where we used the examples at the end of the limit definition page.

Suppose that $\lim_{x \to a} f(x) = 2$ and $\lim_{x \to a} g(x) = 5$. This means that for any tolerance $t > 0$, we have $f(x) \approx 2$ and $g(x) \approx 5$ with tolerance $t$ whenever $x$ is close enough to $a$. We want to show that for any tolerance $t > 0$, we have $f(x)+g(x) \approx 7$ with tolerance $t$ whenever $x$ is close enough to $a$.

At first this seems wrong. For example, consider the tolerance $t=0.1$. If $1.9 < f(x) < 2.1$ and $4.9 < g(x) < 5.1$, we don't necessarily have $6.9 < f(x)+g(x) < 7.1$. If, for example, $f(x)=2.09$ and $g(x)=5.09$, the sum will be $7.18$.

However, we can conclude that $6.8 < f(x)+g(x) < 7.2$ by combining the inequalities: $$ \begin{align} 1.9 &< f(x) < 2.1 \\ 4.9 &< g(x) < 5.1 \end{align} $$

$$ \underbrace{1.9 + 4.9}_{=6.8} < f(x) + g(x) < \underbrace{2.1 + 5.1}_{=7.2} $$ This means that $f(x)+g(x) \approx 7$ with tolerance $0.2$. In general, the tolerance gets multiplied two, making it less precise.

The trick is to make $f(x)$ and $g(x)$ close to their limits within half of the desired tolerance $t$, so that once added, we get the tolerance $t$. For example, $f(x) \approx 2$ with tolerance $0.05$ and $g(x) \approx 5$ with tolerance $0.05$ when $x$ is close enough to $a$. Now we can conclude that $f(x)+g(x) \approx 7$ with tolerance $0.1$.

The same works for any tolerance instead of $0.1$.

Of course, this also works if we replace $2$ and $5$ with any other numbers:

More than two functions¶

If you have more than two functions added, you can think of it as "first function + everything else" and then apply the above two-function version. For example: $$ \begin{align} &~~~~~ \lim_{x \to a} \Bigl(f(x)+g(x)+h(x)+i(x)\Bigr) \\ &= \lim_{x \to a} f(x) + \lim_{x \to a} \Bigl(g(x)+h(x)+i(x)\Bigr) \\ &= \lim_{x \to a} f(x) + \lim_{x \to a} g(x) + \lim_{x \to a} \Bigl(h(x)+i(x)\Bigr) \\ &= \lim_{x \to a} f(x) + \lim_{x \to a} g(x) + \lim_{x \to a} h(x) + \lim_{x \to a} i(x) \\ \end{align} $$ So, if you have any number of things added inside a limit, you can simply take the limit of each thing separately.

Example¶

We calculate $$ \lim_{x \to 2} (4x + 8). $$ In the next derivation, we will look into calculating the limit of a product like $4x$. But we don't know that yet, so let's rewrite it as a sum: $$ 4x = x+x+x+x $$ Now we get $$ \begin{align} \lim_{x \to 2} (4x + 8) &= \lim_{x \to 2} (x+x+x+x + 8) \\ &= \left( \lim_{x \to 2} x \right)+\left( \lim_{x \to 2} x \right)+\left( \lim_{x \to 2} x \right)+\left( \lim_{x \to 2} x \right)+\left( \lim_{x \to 2} 8 \right). \end{align} $$ From these examples, we get $\lim_{x \to 2} x = 2$ and $\lim_{x \to 2} 8 = 8$. Plugging these in gives $$ \lim_{x \to 2} (4x + 8) = 2+2+2+2+8 = 16. $$ We can check this by plugging in a number near 2: $$ 4 \cdot 2.001 + 8 = 16.004 $$ We calculated a limit without plugging in numbers and guessing. This way to calculate limits may seem like more effort, but it's backed by solid mathematical theory, so you can be sure that you didn't miss some surprising behavior when plugging in numbers.

You might notice that simply setting $x=2$ also gives the correct result: $$ 4 \cdot 2 + 8 = 16 $$ This often works, but not always. There is a name for this behavior: we say that a function is continuous if its limits can be calculated this way. We will come back to continuity later.