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Understanding Limits

What is a Limit?

Calculus is the study of change. To understand change, we first need to understand how functions behave as they get incredibly close to a specific point. This is the idea of a limit.

A limit tells us what value a function approaches as its input gets closer and closer to a certain number. The key word here is approaches. We don't actually care what the function's value is at that number, only what it's heading towards.

Imagine walking towards a wall. With every step, you get closer. A limit is like asking: if you could keep taking smaller and smaller steps forever, where would you end up? You would end up right at the wall. The limit is the wall's position, even if you never technically touch it.

We write the limit of a function f(x)f(x) as xx approaches a value cc like this:

limxcf(x)=L\lim_{x \to c} f(x) = L

This equation says, "The limit of f(x)f(x) as xx approaches cc is LL." It means that as the input xx gets arbitrarily close to cc, the output f(x)f(x) gets arbitrarily close to the value LL.

In this graph, the function is not defined at x=2x=2, which is why there's a hole. But as xx gets closer to 2 from either side, the function's value gets closer and closer to 4. Therefore, the limit is 4.

Finding Limits

So how do we find these limit values? Sometimes, it's as simple as it looks.

The easiest method is direct substitution. If the function is well-behaved at the point in question, you can just plug the value in.

For example, let's find the limit of f(x)=x2+5f(x) = x^2 + 5 as xx approaches 3.

limx3(x2+5)\lim_{x \to 3} (x^2 + 5)

We just substitute 3 for xx: 32+5=9+5=143^2 + 5 = 9 + 5 = 14.

The limit is 14. Simple enough. But what happens when direct substitution gives us a problem?

Consider the function from our graph earlier, f(x)=(x24)/(x2)f(x) = (x^2 - 4) / (x - 2). If we try to plug in x=2x=2, we get:

22422=4422=00\frac{2^2 - 4}{2 - 2} = \frac{4 - 4}{2 - 2} = \frac{0}{0}

This result, $0/0$, is called an indeterminate form. It doesn't mean the limit is zero or that it doesn't exist. It's a signal that we need to do more work. We need to use some algebraic techniques.

Factoring is a common technique. We can simplify the function by factoring the numerator and canceling terms.

limx2x24x2=limx2(x2)(x+2)x2\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2}

Since we are only interested in what happens as xx approaches 2, not at x=2x=2, the (x2)(x-2) terms are never zero. So we can safely cancel them.

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

Now we can use direct substitution on the simplified function: $2 + 2 = 4$. This confirms the limit we saw in the graph.

Rationalization is another technique, often used when dealing with square roots.

Let's find the limit:

limx0x+11x\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}

Direct substitution gives 0/00/0. To simplify, we multiply the numerator and denominator by the conjugate of the numerator, which is x+1+1\sqrt{x+1} + 1.

limx0x+11xx+1+1x+1+1\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}

This simplifies the numerator:

limx0(x+1)1x(x+1+1)=limx0xx(x+1+1)\lim_{x \to 0} \frac{(x+1) - 1}{x(\sqrt{x+1} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1} + 1)}

We can cancel the xx terms:

limx01x+1+1\lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1}

Now, using direct substitution:

10+1+1=11+1=12\frac{1}{\sqrt{0+1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}

One-Sided and Infinite Limits

Sometimes a function behaves differently depending on which side you approach a point from. This is where one-sided limits come in.

We use a minus sign for the limit from the left (approaching with values less than cc) and a plus sign for the limit from the right (approaching with values greater than cc).

NotationMeaning
limxcf(x)\lim_{x \to c^-} f(x)Limit as x approaches c from the left.
limxc+f(x)\lim_{x \to c^+} f(x)Limit as x approaches c from the right.

For a regular two-sided limit to exist, the left-hand limit and the right-hand limit must both exist and be equal.

limxcf(x)=L\lim_{x \to c} f(x) = L if and only if limxcf(x)=L\lim_{x \to c^-} f(x) = L and limxc+f(x)=L\lim_{x \to c^+} f(x) = L.

In the graph above, the left and right-hand limits at x=2x=2 are different. Therefore, the overall limit limx2f(x)\lim_{x \to 2} f(x) does not exist.

What if a function grows without bound as it approaches a point? This is called an infinite limit. Consider the function f(x)=1/x2f(x) = 1/x^2 as xx approaches 0.

limx01x2\lim_{x \to 0} \frac{1}{x^2}

As xx gets closer to 0 (from either side), x2x^2 becomes a very tiny positive number. Dividing 1 by a tiny positive number results in a very large positive number. The function is heading towards infinity.

limx01x2=\lim_{x \to 0} \frac{1}{x^2} = \infty

It's important to note that when a limit equals \infty or -\infty, it technically means the limit does not exist, because infinity is not a number. But we write it this way to describe the specific way it doesn't exist: the function is growing without bound.

Continuity

The idea of limits leads directly to the concept of continuity. Informally, a function is continuous if you can draw its graph without lifting your pencil from the paper. There are no breaks, jumps, or holes.

Formally, a function f(x)f(x) is continuous at a point x=cx=c if three conditions are met:

  1. f(c)f(c) is defined (the point exists).
  2. limxcf(x)\lim_{x \to c} f(x) exists (the left and right limits are equal).
  3. limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) (the limit equals the function's value).

If any of these conditions fail, the function has a discontinuity at that point.

Type of DiscontinuityDescriptionExample Graph
RemovableA "hole" in the graph. The limit exists, but it's not equal to the function's value (or the function isn't defined there).Our first graph of f(x)=(x24)/(x2)f(x) = (x^2-4)/(x-2).
JumpThe graph "jumps" from one value to another. The left and right-hand limits exist but are not equal.Our one-sided limit step function graph.
InfiniteThe function goes to infinity at a vertical asymptote.The graph of f(x)=1/x2f(x)=1/x^2 has an infinite discontinuity at x=0x=0.

Understanding limits and continuity is the first major step in calculus. It sets the stage for defining the rate of change (derivatives) and the accumulation of quantities (integrals).