# Limits

Definition: Calculus is often defined as “The Study of Limits”, so in order to understand calculus, it is important to first understand what a limit is. Let’s start with a really technical definition, and then talk about what it means. Say you have a function, $f(x)$, and an x-value, $x = a$. We define “the limit of f(x) as x approaches a, written $\lim\limits_{x \rightarrow a} f(x)$, to be L if:

For all $\epsilon > 0$, there exists a $\delta > 0$ such that, if $|x - a| < \delta$ then $|f(x) - f(a)| < \epsilon$

That’s certainly a mouthful! But what does it mean? Well, simply put, the limit of a function f(x) at a point x=a is simply what happens to the value of the function as x gets very, very close to a. Take a look at this image:

As you can see, as long as we pick an x that is very close to a, our function will be very close to L. If we want to get closer to L, we can just pick a smaller δ; the closer x gets to a, the closer f(x) gets to L.

So, how can we find a limit? There are a few methods you can try. The first is simply called plugging in, and all you do is plug the value of a into the functions. For many functions, .

Example: Find the limit as x approaches 5 of $f(x) = x^2 + 3x - 7$.

Solution:

Here, plugging in will work:

Another useful technique is factoring, which is useful for rational functions.

Example: Find the limit as x approaches 1 of  .

Solution:

Let’s try plugging in again:

Unfortunately, it didn’t work! This function is undefined at x=1, so we can’t simply plug in to evaluate the limit. However, we can factor and cancel:

This technique works really well for functions that are fractions of polynomials. For fractions involving square roots, we’ll use a similar method called multiplying by the conjugate.

Example: Find the limit of  as x approaches 9.

Solution:  We can see very quickly that plugging in won’t work; we’ll get a 0 in the denominator again. But the numerator has this expression: $3 + \sqrt{x}$, so we can try to simplify by multiplying by its conjugate.

Recall that the conjugate of a a radical expression switches the sign of the square root piece; here, the conjugate of $3 - \sqrt{x}$ is $3 + \sqrt{x}$. When you multiply a radical expression by its conjugate, it will be rational; that is, the square root will go away. In order to not change the fraction, we have to multiply both the top and the bottom by the same number:

As you can see, multiplying by the conjugate allowed us to cancel the factor from the denominator that was giving us a 0, allowing us to plug into the function to obtain the result.