Question

1. determine the point(s), if any, at which the function is discontinuous. classify any discontinuity as jump, removable, infinite, or other.

(i) $\frac{x + 1}{x^{2}+1}$ (ii) $\frac{x^{2}-9}{x^{2}-5x + 6}$ (iii) $\frac{|x - 3|}{x - 3}$ (iv) $\tan(2x)$

Explanation:

Step1: Analyze (i) $\frac{x + 1}{x^{2}+1}$

The denominator $x^{2}+1$ is always positive for all real - valued $x$ since $x^{2}\geq0$ for all $x\in R$, so $x^{2}+1\geq1$. Thus, the function is continuous for all $x\in R$.

Step2: Analyze (ii) $\frac{x^{2}-9}{x^{2}-5x + 6}$

Factor the numerator and denominator: $x^{2}-9=(x + 3)(x - 3)$ and $x^{2}-5x + 6=(x - 2)(x - 3)$. The function is undefined at $x = 2$ and $x=3$.
$\lim_{x
ightarrow2}\frac{(x + 3)(x - 3)}{(x - 2)(x - 3)}=\lim_{x
ightarrow2}\frac{x + 3}{x - 2}=\infty$, so there is an infinite discontinuity at $x = 2$.
$\lim_{x
ightarrow3}\frac{(x + 3)(x - 3)}{(x - 2)(x - 3)}=\lim_{x
ightarrow3}\frac{x + 3}{x - 2}=6$, so there is a removable discontinuity at $x = 3$.

Step3: Analyze (iii) $\frac{|x - 3|}{x - 3}$

We know that $|x - 3|=

$$\begin{cases}x - 3, & x\geq3\\-(x - 3), & x<3\end{cases}$$

$.
$\lim_{x
ightarrow3^{+}}\frac{|x - 3|}{x - 3}=\lim_{x
ightarrow3^{+}}\frac{x - 3}{x - 3}=1$ and $\lim_{x
ightarrow3^{-}}\frac{|x - 3|}{x - 3}=\lim_{x
ightarrow3^{-}}\frac{-(x - 3)}{x - 3}=-1$. So there is a jump discontinuity at $x = 3$.

Step4: Analyze (iv) $\tan(2x)$

The tangent function $y = \tan(t)$ is discontinuous when $t=(2n + 1)\frac{\pi}{2},n\in Z$. For $y=\tan(2x)$, we set $2x=(2n + 1)\frac{\pi}{2}$, then $x=(2n + 1)\frac{\pi}{4},n\in Z$. At these points, $\lim_{x
ightarrow(2n + 1)\frac{\pi}{4}^{+}}\tan(2x)=\infty$ and $\lim_{x
ightarrow(2n + 1)\frac{\pi}{4}^{-}}\tan(2x)=-\infty$, so there are infinite discontinuities at $x=(2n + 1)\frac{\pi}{4},n\in Z$.

Answer:

(i) No discontinuities.
(ii) Infinite discontinuity at $x = 2$, removable discontinuity at $x = 3$.
(iii) Jump discontinuity at $x = 3$.
(iv) Infinite discontinuities at $x=(2n + 1)\frac{\pi}{4},n\in Z$.