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.


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  limit4.


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 limit7 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.

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