  ## Exercises - Continuous Functions (and Review)

1. Determine where $f(x)=\sqrt{x}$ is continuous.

$\sqrt{x}$ is an even root function. Recall even root functions (i.e., $\sqrt[n]{x}$ for even $n \gt 1$) are continuous on $(0,\infty)$. They are also continuous from the right at $x=0$.

2. Determine where $\displaystyle{f(x) = \sin \left( \frac{1}{x} \right)}$ is continuous.

Note that $1/x$ is a rational function and continuous on its domain (i.e., $x \ne 0$). The function $\sin x$ is also continuous on its domain (i.e, all reals). Recalling that if a function $g$ is continuous at $x=c$ and a function $f$ is continuous at $x=g(c)$, then the composition $f(g(x))$ is continuous at $c$, we can conclude the composition $\sin (1/x)$ is continuous for all $x \ne 0$. At $x=0$, however, $\sin (1/x)$ must have a discontinuity -- for two reasons. First, this function is not defined at $x=0$. Second, we have previously seen this function has infinite (non-damped) oscillatory behavior at this $x$-value. Thus, the limit of $\sin (1/x)$ as $x \rightarrow 0$ does not exist -- which also keeps this function from being continuous at $x=0$.

3. Determine where $\displaystyle{f(x) = x \sin \left( \frac{1}{x} \right)}$ is continuous.

As $1/x$ is continuous on its domain (i.e,. $x \ne 0$) being a rational function, and $\sin x$ is continuous on its domain (all reals), their composition $\sin (1/x)$ is continuous for all $x \ne 0$. Further, the function $x$ itself is continuous everywhere, being a polynomial function. Thus, the product of these two, $x \sin (1/x)$ is also continuous for all $x \ne 0$. The question of continuity at $x=0$ remains. As the $x \sin (1/x)$ is undefined at this $x$-value, it must be the case that it is not continuous at $x=0$.

4. Discuss the discontinuities of the following function:

$$f(x) = \left\{ \begin{array}{cc} \cos(2x), & x \le -\pi\\ -x, & -\pi \lt x \lt 0\\ x^2, & 0 \lt x \lt 2\\ \displaystyle{\frac{x-4}{x-3}}, & x \ge 2 \end{array} \right.$$

Since $f(x)$ is a piecewise-defined function, as we hunt for discontinuities, we need to be alert to two different things:

1. relevant discontinuities inside the "pieces" that make up the function; and
2. if there are any discontinuities corresponding to the "borders between pieces"

Note that the first three pieces are continuous everywhere. The first is a composition of continuous functions, while the next two are polynomials.

The fourth piece, however, has a discontinuity due to a domain issue at $x=3$. It is important to check if this is a "relevant" discontinuity. Note that $x = 3$ is one of the $x$-values associated with this "piece" (i.e., $x \ge 2$). Had this not been the case, the domain issue would never have actually been realized -- and thus there would likely not have been a discontinuity there.

Knowing there is a discontinuity at $x=3$, we find the corresponding limit so that we can graphically interpret what kind of discontinuity happens at this value:

$$\lim_{x \rightarrow 3^-} f(x) = \lim_{x \rightarrow 3^-} \frac{x-4}{x-3} = +\infty \quad \textrm{(Note, this is a }\frac{-1}{0^-} \textrm{ situation)}$$ $$\lim_{x \rightarrow 3^+} f(x) =\lim_{x \rightarrow 3^+} \frac{x-4}{x-3} = -\infty \quad \textrm{(Note, this is a }\frac{-1}{0^+} \textrm{ situation)}$$

Thus, at $x=3$ we have an non-removable (infinite) discontinuity due to the presence of a vertical asymptote there.

The second place we said we needed to look was at the "borders between pieces". This is due to the fact that the behavior to the left and right of these border values often differs (they correspond to different pieces, after all). This can result in different limiting behaviors on the left and right of these border values, which can affect the continuity there.

We must also be alert to domain issues at these borders, should they exist, as this too can affect continuity.

Here, we have three "border" values to check: $x = -\pi$, $0$, and $2$. We consider each of these in turn.

At $x=-\pi$:

$$\lim_{x \rightarrow -\pi^-} f(x) = \lim_{x \rightarrow -\pi^-} \cos(2x) = 1$$ $$f(-\pi) = \cos(2\pi) = 1$$ $$\lim_{x \rightarrow -\pi^+} f(x) = \lim_{x \rightarrow -\pi^+} -x = \pi$$

As the left and right limits disagree, we have a non-removable gap discontinuity at $x=-\pi$ from $(-\pi,1)$ to $(-\pi,\pi)$.

At $x=0$:

$$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} -x = 0$$ $$f(0) \textrm{ is undefined (i.e., no piece addresses it)}$$ $$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} x^2 = 0$$

Here, the left and right limits agree, meaning $\lim_{x \rightarrow 0} f(x) = 0$.

However, this in combination with noticing $f(0)$ is undefined tells us we have a removable discontinuity at $x=0$ taking the form of a hole at $(0,0)$.

At $x=2$:

$$\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} x^2 = 4$$ $$f(2) = \frac{2-4}{2-3} = 2$$ $$\lim_{x \rightarrow 2^+} f(x) = \lim_{x \rightarrow 2^+} \frac{x-4}{x-3} = \frac{2-4}{2-3} = 2$$

As the left and right limits disagree, we have a non-removable gap discontinuity at $x = 2$ from $(2,4)$ to $(2,2)$.

5. Discuss the discontinuities of the following function:

$$f(x) = \left\{ \begin{array}{cc} \displaystyle{\frac{2x+3}{x+2}}, & x \lt 1\\ x^2, & 1 \le x \lt 2\\ 5x-6, & x \ge 2 \end{array} \right.$$

There are two discontinuities; at $x = 1$ and $x = -2$.

The discontinuity at $x=1$ is a result of $\lim_{x \rightarrow 1} f(x)$ failing to exist as its left and right limits disagree ($\lim_{x \rightarrow 1^-} f(x) = 5/3$ while $\lim_{x \rightarrow 1^+} f(x) = 1$). This means there is a non-removable gap discontinuity at $x=1$ from $(1,5/3)$ to $(1,1)$.

The discontinuity at $x=-2$ occurs for two reasons. First, $f(-2)$ is undefined. Second, $\lim_{x \rightarrow -2} f(x)$ does not exist (it is $+\infty$ on the left, and $-\infty$ on the right). This means there is a non-removable discontinuity at $x=-2$ that takes the form of a vertical asymptote at this $x$-value.

6. Suppose $\displaystyle{\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = a}$ and $\displaystyle{\lim_{x \rightarrow c} \frac{g(x)}{h(x)} = b}$. Find $\displaystyle{\lim_{x \rightarrow c} \frac{f(x)}{h(x)}}$; justify your response.

We can find this limit by first multiplying by a "well chosen value of $1$" and then applying the limit law for a product of limits, as shown below.

$\displaystyle{\lim_{x \rightarrow c} \frac{f(x)}{h(x)} = \lim_{x \rightarrow c} \frac{f(x)}{h(x)} \cdot \frac{g(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f(x)}{g(x)} \cdot \frac{g(x)}{h(x)} = \left(\lim_{x \rightarrow c} \frac{f(x)}{g(x)} \right) \left( \lim_{x \rightarrow c} \frac{g(x)}{h(x)} \right) = ab}$

7. Let $f(x)$ and $g(x)$ be two functions that both approach $\pm \infty$ as $x \rightarrow c$. We say that $f(x)$ has a higher order of growth than $g(x)$ if $\displaystyle{\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \pm \infty}$. Rank the following functions in increasing order of growth as $x \rightarrow +\infty$; $f(x) = x$, $g(x) = \sqrt{x}$, $h(x) = (x+1)^2$.

$\sqrt{x}$, $x$, $(x+1)^2$