# Suppose g(x) = \frac{1}{x - 2} if x < 1, and 2x - 4 if x is greater than or equal to 1. The best...

## Question:

Suppose {eq}\; g(x) = \left\{\begin{matrix} \displaystyle\frac{1}{x - 2} & \textrm{ if } x < 1 \\ 2x - 4 **** \textrm{ if } x \geq 1 \end{matrix}\right. {/eq}

The best description concerning the continuity of {eq}g(x) {/eq} is that the function

A.) {eq}\textrm{ is continuous.} {/eq}

B.) {eq}\textrm{ has a jump discontinuity.} {/eq}

C.) {eq}\textrm{ has an infinite discontinuity.} {/eq}

D.) {eq}\textrm{ has a removable discontinuity.} {/eq}

E.) {eq}\textrm{ None of the above.} {/eq}

## Continuity:

We are given a function and asked to determine the continuity. To do this, we have to make sure we account for every point and inspect any possible discontinuities. This is true for piecewise-defined functions in particular since even if the functions that are given are continuous, discontinuities could exist at the points where the function changes definition.

We are asked to inspect the continuity of the function $$g(x) = \left\{ \begin{array}{rl} \dfrac{1}{x - 2} & \text{if } x < 1 \\ 2x - 4 & \text{if } x \geq 1 \end{array} \right.$$

Note that for the given intervals, each of the functions in the definition is continuous because {eq}\dfrac{1}{x - 2} {/eq} is continuous everywhere except {eq}x = 2 {/eq} which is not contained in the interval where this function is given. Also, {eq}2x - 4 {/eq} is a polynomial so it is continuous everywhere. The only possible discontinuity is where the function changes its definition, which is at {eq}x = 1 {/eq}.

We can find the limit from both sides to determine the continuity at this point. We have \begin{align*} \lim_{x \to 1^-}{g(x)} &= \lim_{x \to 1^-}{\frac{1}{x - 2}} \\ &= -1 \end{align*} and \begin{align*} \lim_{x \to 1^+}{g(x)} &= \lim_{x \to 1^+}{(2x - 4)} \\ &= -2 \end{align*} and since the limits are different, we can say that {eq}g {/eq} has a jump discontinuity.