Finding Limits

Take \(f(x)=k\) where \(a\) and \(k\) are constant real numbers. Is it true that

\[lim_{x \rightarrow a} f(x)=k\]

Answer:

Yes. Using the definition, for any \(\epsilon > 0\) you are given,

\[|f(x)-k|=|k-k|=0< \epsilon\]

no matter what \(\delta\) you use. So constant functions have the limit you would expect them to.

Take \(f(x)=x\), and let \(a\) be a constant real number. How do you know that

\[lim_{x \rightarrow a} f(x)=a\]

Answer:

You might be tempted to say "of course, the limit is \(a\) - the function is just a line". In fact, that is almost enough. You can't use any of the Properties of Limits, but you can use the definition and take \(\delta = \epsilon\) to show that the limit is \(a\).

Take the function \(f(x)=2x^3-3x^2+7\), and \(a\) to be a constant real number. Find

\[lim_{x \rightarrow a} f(x)\]

Answer:

Notice that the function is just the sum and product of powers of \(x\) along with the constant \(7\). You already know that

\(lim_{x \rightarrow a} x=a\) and \(lim_{x \rightarrow a} 7 =7\)

from the two examples above, which means the conditions to apply the Sum Rule, Product Rule, and Constant Rule are met. Then applying them gives

\[lim_{x \rightarrow a} f(x)= lim_{x \rightarrow } ( 2x^3-3x^2+7)\]

\[lim_{x \rightarrow a} f(x)=2a^3-3a^2+7\]

Consider the function

\[f(x)=\dfrac{1}{4}(x+1)(x-1)(x-5)\]

Find the limit of the function as \(x \rightarrow 3 \).

Answer:

First, graph the function and make a table of values near \(x=3\). Although the function has more roots than are shown in the graph, since you only care about the limit as \(x \rightarrow 3\), it makes sense to zoom in on the function there.

Using a graph with multiple points to find the limit of a function in red.

Table 1. Limit example points.

The points on the graph correspond to the points in the table. You can see from both the graph and table that as \(x\) gets closer and closer to \(x= 3\), the function values get closer and closer to \(-4\) That means that

\[lim_{x \rightarrow 3} f(x)=-4\]

.

Notice that you don't actually care about the function value at \(x=3\) when finding the limit, because the definition says to look close to \(x=3\) but not at \(x=3\).

From the previous example, we had

\[f(x)=\dfrac{1}{4}(x+1)(x-1)(x-5)\]

and found the limit as \(x \rightarrow 3\). Since you know that all polynomials are continuous everywhere (see Continuity and see Theorems of Continuity for more details), you know the limit of the function exists and is equal to the function value. Since \(f(3)=-4\), that means

\[lim_{x \rightarrow 3} =-4\]

Look at the function

\[f(x)=\dfrac{x^2-2x-8}{x-4}\]

and find the limit as \(x \rightarrow 4\).

Answer:

The function is undefined at \(x=4\), so you can't just plug in the function value to find the limit. But you can factor the numerator to get

\[f(x)=\dfrac{x^2-2x-8}{x-4}=\dfrac{(x-4)(x+2)}{x-4}=x+2\]

as long as \(x \neq 4\). That means the graph of the function is actually the straight line \(y=x+2\) with a hole at the point \((4, 6)\). So \(lim_{x \rightarrow 4} f(x)=6\).