Chapter 2.3, Problem 40E

To determine

To find: The reason for the continuity of the function $y=\{\begin{array}{l}\frac{x^2-1}{x-1},x\ne 1\\ 2,x=1\end{array}$ , check the domain of the function, also mention the theorems used in the continuity of function with the functions assumed as continuous.

Answer to Problem 40E

The domain of the function $y=\{\begin{array}{l}\frac{x^2-1}{x-1},x\ne 1\\ 2,x=1\end{array}$ is equal to interval $\left(-\infty ,\infty \right)$ and the difference and quotient theorems are used and the functions $x$ and $1$ are assumed as continuous.

Explanation of Solution

Given information:

The function is $y=\{\begin{array}{l}\frac{x^2-1}{x-1},x\ne 1\\ 2,x=1\end{array}$ .

Calculation:

The function corresponding to the value $x\ne 1$ is $y=\frac{x^2-1}{x-1}$ .

Assume the function $x$ as a continuous function.

The function $y=\frac{x^2-1}{x-1}$ is the quotient of two functions $h\left(x\right)=x^2-1$ and $j\left(x\right)=x-1$ .

The function $h\left(x\right)=x^2-1$ is the subtraction of functions $x^2$ and $1$ .

The function $x^2$ is the product of $x$ twice. So, by the product theorem the function $x^2$ is also continuous function.

The function $h\left(x\right)=x^2-1$ is the subtraction of continuous functions $x^2$ and $1$ . So, the function $h\left(x\right)=x^2-1$ is also a continuous function.

The function $j\left(x\right)=x-1$ is the subtraction of continuous functions $x$ and $1$ . So, the function $j\left(x\right)=x-1$ is also a continuous function.

As the function $y=\frac{x^2-1}{x-1}$ is the quotient of two continuous functions $h\left(x\right)=x^2-1$ and $j\left(x\right)=x-1$ . So, by quotient theorem the function $y=\frac{x^2-1}{x-1}$ is continuous function.

The value of $y$ at $x=1$ is $2$ .

The function $y=\{\begin{array}{l}\frac{x^2-1}{x-1},x\ne 1\\ 2,x=1\end{array}$ is defined for all the real values of $x$ . So, the domain of the function $y=\{\begin{array}{l}\frac{x^2-1}{x-1},x\ne 1\\ 2,x=1\end{array}$ is equal to interval $\left(-\infty ,\infty \right)$ .

Therefore, the domain of the function $y=\{\begin{array}{l}\frac{x^2-1}{x-1},x\ne 1\\ 2,x=1\end{array}$ is equal to interval $\left(-\infty ,\infty \right)$ and the difference and quotient theorems are used and the functions $x$ and $1$ are assumed as continuous.