# Limit of Composite Functions

Recently, a number of questions about the limit of composite functions have been discussed on the AP Calculus Community bulletin board and also on the AP Calc TEACHERS – AB/BC Facebook page. The theorem that we would like to apply in these cases is this:

If f is continuous at b and $\underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)=b$, then $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( {g\left( x \right)} \right)=f\left( b \right)$.

That is, $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( {g\left( x \right)} \right)=f\left( {\underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)} \right)$

The problem is that in the examples one or the other of the hypotheses (continuity or the existence of  $\underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)$) is not met. Therefore, the theorem cannot be used. This does not mean that the limits necessarily do not exist, rather that we need to find some other way of determining them. We need a workaround. Let’s look at some.

Example 1: The first example is from the 2016 BC International exam, question 88. Students were given the graph at the right and asked to find  $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {1-{{x}^{2}}} \right)$.

At first glance it appears that as x approaches zero,  $\left( {1-{{x}^{2}}} \right)$ approaches 1 and the limit does not exist since f is not continuous at 1, so the theorem cannot be used. However, on closer examination, we see that $\left( {1-{{x}^{2}}} \right)$ is always less than 1, so $\left( {1-{{x}^{2}}} \right)$ is approaching 1 from the left (or from below). Therefore, as f approaches 1 from the left $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {1-{{x}^{2}}} \right)=\underset{{x\to {{1}^{-}}}}{\mathop{{\lim }}}\,f\left( x \right)=3$

Another approach is to try to write the equation of f. Although we cannot be certain, it appears that: $f\left( x \right)={{x}^{2}}+2,x<1$.

Then,  $f\left( {1-{{x}^{2}}} \right)={{\left( {1-{{x}^{2}}} \right)}^{2}}+2={{x}^{4}}-2{{x}^{2}}+3,x<1$. In this form the limit is obviously 3.

Example 2: The second example is also based on a graph. Given the graph of a function f, shown at the left, what is  $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)$ ?

$\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)=2$. Since f is not continuous at 2, the theorem cannot be used. But, notice that as x approaches 0 from both sides, the limit 2 is approached from the left (from below). So we need to find the value of f as its argument approaches  ${{2}^{-}}$. From the graph, this value is zero; So, $\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)=0$

To clarify this a little more, let’s look at a similar problem suggested by Sondra Edwards on the Facebook site: Consider this similar function:

Now as we approach 0 from both sides  $\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)=2$ approached from both sides. But now f(2) does not exist (DNE). (This is the “outside” f, which is not continuous here.) This time, the limiting value, 2, is approached from both sides. Therefore, $\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)$ DNE. There is no way to work around the discontinuity.

For a similar question see here

Example 3: If $\displaystyle f\left( x \right)=\frac{1}{x},x\ne 0$, what is $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( {x)} \right)} \right)$ ?

Here again, the theorem cannot be used, since the “inside” function has no limit as x approaches 0. But, this function is its own inverse, so $\displaystyle f\left( {f\left( {x)} \right)} \right)=x$, and $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)=\underset{{x\to 0}}{\mathop{{\lim }}}\,x=0$

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