Q. 19 - Calculus

Question

Graph the functions $f\left(x\right)=x+1$ and $g\left(x\right)=\frac{x^2-1}{x-1}$, and show that they are equal everywhere except at one point. Then show that f(x) and g(x) have different values, but the same limit, at this point. 

Step-by-Step Solution

Answer

Both functions approach 2 when x approaches 1. 

Step 1. Given information.

Given functions are:

$$\begin{gathered} f\left(x\right)=x+1 \\ g\left(x\right)=\frac{x^2-1}{x-1} \end{gathered}$$

We have to graph the function.

Step 2. Graph the f(x)

Graph of $f\left(x\right)=x+1$ is:

Step 3. Graph the g(x)

Graph of $g\left(x\right)=\frac{x^2-1}{x-1}$ is:

Step 4. Compare the graphs.

The graph of g looks similar to the graph of f but with the hole at $x=1$

Here,

$f\left(1\right)=2$ but $g\left(1\right)$ is undefined.

and both functions approach 2 when x approaches 1.