Calculus

Introduction with Limits

Limits are the basis for almost everything in calculus. By the time you get to derivatives, you'll be doing limits and you won't even know it. To become familiar with what a limit is, let's consider the function

f(x)=x21x1.f(x) = \frac{x^2-1}{x-1}.
Try this.

What is the value of f(x)f(x) at x=1x=1?

Hint 1

Don't be scared! Just plug it in!

Solution 1
12111=00.\frac{1^2 - 1}{1-1} = \frac{0}{0}.

However, we can't divide by 00. So, f(x)f(x) must not be defined at x=1x=1.

00\frac{0}{0} is indeterminate, so the function f(x)f(x) would be indeterminate at x=1x=1. However, the function is defined at x=0.9x=0.9 and x=1.1x=1.1:

f(0.9)=1.9f(0.9) = 1.9
f(1.1)=2.1f(1.1) = 2.1

In fact, the function is defined for any 1±δ,1\pm \delta, where δ0.\delta \neq 0.

In this lesson, we'll be considering the case where δ\delta is very close, but not equal to 0.0.

Try dragging the black point to x=1x=1 in the above graph, and you will notice that the point disappears.

However, notice that it "looks like"

f(1)=2.f(1)=2.

So, we can say that

limx1f(x)=2.\lim_{x\to 1} f(x) = 2.

The English for this is "As xx approaches 1, f(x)f(x) approaches 22."

Another possible interpretation is "The limit of f(x)f(x) as xx approaches 1 is 22."

This does not mean that f(1)=2f(1)=2 , but rather that if xx is very close to 11, then is f(x)f(x) very close to 22.

Look closely at the graph.

Notice that the graphed function is linear. What linear function does it look like?

Hint 1

What's the slope? The y-intercept?

Remember,

y=mx+b.y=mx+b.
Solution 1

We can find two points, (0,1)(0,1) and (2,3)(2,3) (remember, we can't use (1,2)(1,2)!), and use these to find the slope.

m=3120=22=1.m=\frac{3-1}{2-0} = \frac{2}{2} = 1.

Then, we can see the y-intercept is at (0,1)(0,1). Our slope-intercept equation is

y=x+1.y=x+1.

The function covers almost all points covered by the function x+1x+1.

We can confirm this by performing the following operation:

x21x1=(x1)(x+1)x1=x+1\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1

Important! This kind of math is only allowed in limits. Notice how we divided the top and bottom of the fraction by x1x-1, a non-constant – we only did this knowing that xx is very close to, but not equal to 1.

Now try this:

limx2x25x+6x2.\lim_{x\to 2} \frac{x^2-5x+6}{x-2}.

As we said before, it is okay to divide by non-constants in limits. So, we can say

limx2x25x+6x2=limx2x3.\lim_{x\to 2} \frac{x^2-5x+6}{x-2} = \lim_{x\to 2} x-3.

We can just plug in x=2x=2 here to find

limx2x25x+6x2=1.\lim_{x\to 2} \frac{x^2-5x+6}{x-2} = -1.
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