No history yet

Limits

Approaching a Point

Calculus is the study of change, and at its very heart is the idea of a limit. A limit answers a simple question: what value is a function approaching as its input gets closer and closer to a certain number? Notice the word "approaching." We're interested in the journey, not necessarily the final destination.

Let's look at the function f(x)=x24x2f(x) = \frac{x^2 - 4}{x - 2}. If we try to find the value of this function at x=2x = 2, we run into a problem. Plugging in 2 gives us 4422=00\frac{4 - 4}{2 - 2} = \frac{0}{0}, which is undefined. The function simply doesn't exist at x=2x = 2. But what happens as we get really close to 2?

xf(x) = (x²-4)/(x-2)
1.93.9
1.993.99
1.9993.999
??
2.0014.001
2.014.01
2.14.1

As you can see from the table, the closer our input xx gets to 2, the closer the output f(x)f(x) gets to 4. It doesn't matter that the function is undefined right at 2. The limit is 4. We write this formally as:

limx2x24x2=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4

This reads as "the limit of the function as xx approaches 2 is 4."

Techniques for Finding Limits

For many simple functions, you can find the limit by just plugging the number in. The limit of f(x)=x2+1f(x) = x^2 + 1 as xx approaches 3 is just 32+1=103^2 + 1 = 10. But as we've seen, that doesn't always work.

When direct substitution gives you an "indeterminate form" like 00\frac{0}{0}, you need to use other techniques to find the limit.

One common technique is factoring. Let's revisit our original function. We can factor the numerator:

f(x)=x24x2=(x2)(x+2)x2f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}

As long as xx isn't exactly 2, we can cancel the (x2)(x-2) terms. Since a limit only cares about what happens as we approach 2, this is perfectly fine. We get a simplified function, f(x)=x+2f(x) = x+2. Now, finding the limit is easy:

limx2(x+2)=2+2=4\lim_{x \to 2} (x+2) = 2+2=4

This confirms our answer from the table.

One-Sided Limits

Sometimes a function behaves differently depending on whether you approach a number from the left (smaller values) or the right (larger values). These are called one-sided limits.

The limit as xx approaches cc from the left is written with a minus sign: limxcf(x)\lim_{x \to c^-} f(x).

The limit as xx approaches cc from the right is written with a plus sign: limxc+f(x)\lim_{x \to c^+} f(x).

For a general limit to exist, the left-hand limit must equal the right-hand limit.

In the graph above, as xx approaches 1 from the left, the function gets closer to a yy-value of 2. But as xx approaches 1 from the right, the function gets closer to 3. Since 232 \neq 3, the overall limit limx1f(x)\lim_{x \to 1} f(x) does not exist.

Limits at Infinity and L'Hôpital's Rule

We can also ask what happens to a function as xx gets infinitely large, either positive (xx \to \infty) or negative (xx \to -\infty). This tells us about the function's long-term behavior. For example, consider f(x)=1/xf(x) = 1/x. As xx becomes huge, 1/x1/x gets incredibly small, approaching 0. So, limx1/x=0\lim_{x \to \infty} 1/x = 0.

For rational functions (a polynomial divided by a polynomial), a quick trick is to divide every term by the highest power of xx in the denominator. This often simplifies the problem dramatically.

Finally, there's a powerful tool for tricky limits called L'Hôpital's Rule. It's used for indeterminate forms like 00\frac{0}{0} or \frac{\infty}{\infty}. The rule states that if you have such a limit, you can take the derivative of the numerator and the derivative of the denominator separately, and then try to find the limit again.

L'Hôpital's Rule: If limxcf(x)g(x)\lim_{x \to c} \frac{f(x)}{g(x)} results in 00\frac{0}{0} or \frac{\infty}{\infty}, then limxcf(x)g(x)=limxcf(x)g(x)\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}.

Let's try it on a new example: limx0sin(x)x\lim_{x \to 0} \frac{\sin(x)}{x}. Plugging in 0 gives sin(0)0=00\frac{\sin(0)}{0} = \frac{0}{0}. Time for L'Hôpital's Rule.

The derivative of sin(x)\sin(x) is cos(x)\cos(x). The derivative of xx is 11.

So, our new limit is:

limx0cos(x)1=cos(0)1=11=1\lim_{x \to 0} \frac{\cos(x)}{1} = \frac{\cos(0)}{1} = \frac{1}{1} = 1

L'Hôpital's Rule provides an elegant way to solve limits that would otherwise be very difficult.

Quiz Questions 1/6

What does the limit of a function, limxcf(x)\text{lim}_{x \to c} f(x), describe?

Quiz Questions 2/6

Evaluate the following limit: limx3x29x3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Understanding limits is the first major step into the world of calculus. They form the foundation for everything that follows.