Uniqueness of Limit¶

Can we have $\lim_{x \to 2} f(x) = 5$ and $\lim_{x \to 2} f(x) = 8$ for the same function $f$?

If you think of the notation such that $\lim_{x \to 2} f(x)$ by itself is a thing, then this is obviously not possible; there's no way how something could be equal to $5$ and $8$ at the same time. However, that's not how we have defined the limit. The definition doesn't say anything about what $\lim_{x \to 2} f(x)$ by itself is; it only tells what setting that equal to something means, such as $\lim_{x \to 2} f(x) = 5$.

Suppose $\lim_{x \to 2} f(x) = 5$ and $\lim_{x \to 2} f(x) = 8$ for the same function $f$. Let's choose the tolerance $t=0.1$. If $x$ is close enough to $2$, we have $f(x) \approx 5$ with tolerance $0.1$, so $f(x) < 5.1$. Also, if $x$ is close enough to $2$, we have $f(x) \approx 8$ with tolerance $0.1$, so $f(x) > 7.9$. This is not possible: no number can be less than $5.1$ and greater than $7.9$. This means that a limit cannot simultaneously be $5$ and $8$.

The same applies for any two different numbers instead of $5$ and $8$ as long as we pick a small enough tolerance.

Also, the limit can't be equal to more than two different numbers, because if a limit was equal to more than two different numbers, then we could pick two of those numbers; the limit would be equal to them, and that's impossible by the above result.

An important consequence of this result is that if there is some number that the limit equals to (in other words, if the limit exists, unlike in this example, then we can write things like $$ \left( \lim_{x \to a} f(x) \right)^2 + 7. $$ This is a shorthand for letting $L$ be the only number for which $\lim_{x \to a} f(x) = L$, and then writing $L^2+7$. We can think of $\lim_{x \to a} f(x)$ as a number, because we know that it can't be equal to multiple different numbers.