- Show that \(f(x)=\frac{x^2+3}{x+7}\) is continuous at the point \(x=1\).
This is just to practice using the Continuity Test.
1. \(\lim_{x\rightarrow 1} x^2+3=1+3=4\) since this is a polynomial, and similarly \(\lim_{x\rightarrow 1} x+7=1+7=8\). Now by the quotient rule of the limit laws, \(\lim_{x\rightarrow 1} f(x)=\frac{4}{8}=\frac{1}{2}\).
2. The denominator of \(f\) is not equal to zero at \(x=1\), so \(f(1)\) is defined.
3. \(f(1)=\frac{1+3}{1+7}=\frac{1}{2}=\lim_{x\rightarrow 1}f(x)\).
Since all three criteria are satisfied, this shows \(f(x)\) is continuous at \(x=1\).
- Come up with an example of a problem whose limit exists at \(x=8\), is defined at \(x=8\) but is not continuous at \(x=8\).
For example,
\[
f(x)=
\begin{cases}
x, & x \neq 8\\
0, & x=8.
\end{cases}
\]
You can check that \(\lim_{x\rightarrow 8}f(x)=8\), and \(f(8)=0\). The first two criteria of the Continuity Test are satisfied, but the third is not.
- Show that the equation \(\cos x=x\) has a solution between 0 and 1.
Consider the function \(h(x)=\cos x-x\). The problem is equivalent to showing that \(h\) has a root between 0 & 1.
\(h\) is a continuous function on [0,1] since it is the difference of functions which are continuous on [0,1].
(You should note that both sine and cosine are continuous on \((-\infty,\infty)\).) Now, \(h(0)=1-0=1>0\) and \(h(1)=\cos(1)-1<0\). The latter fact should be clear because \(\cos(0)=1\) and then starts decreasing as we consider values greater than 0, but remains positive through \(x=\pi/2\). Thus, by the Intermediate Value Theorem, since \(h\) is continuous on [0,1], it must be that \(h(x)=0\) for some \(x\in [0,1]\), which implies \(\cos x=x\) for some \(x\in [0,1]\).