Question

Show that the function ${\displaystyle \{\begin{array}{ll}\frac{x^3-1}{x-1}\text{,}& \text{if} x\ne 1\\ 3\text{,}& \text{if} x=1\end{array}}$ is continuous om ${\displaystyle (-\infty ,\infty )}$

Answer 1

${\displaystyle \{\begin{array}{ll}\frac{x^3-1}{x-1}\text{,}& \text{if} x\ne 1\\ 3\text{,}& \text{if} x=1\end{array}}$

Clearly, the given function f(x) is defined at all points of R.

Case (i) Let x₀ ∈ ${\displaystyle (-\infty ,1)}$ then

${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)= \underset{x\to x_0}{lim}\frac{x^3-1}{x-1}}$

= ${\displaystyle \frac{x_0^3-1}{x_0-1} x_0\ne 1}$

${\displaystyle f\left(x_0\right)=\frac{x_0^3-1}{x_0-1} x_0\ne 1}$

∴ ${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)=f\left(x_0\right)}$

f(x) is continuous at x = x₀.
Since x₀ is arbitrary f(x) is continuous at all points of ${\displaystyle (-\infty ,1)}$.

Case (ii) Let x₀ ∈ ${\displaystyle (1,\infty )}$ then

${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)= \underset{x\to x_0}{lim}\frac{x^3-1}{x-1}}$

= ${\displaystyle \frac{x_0^3-1}{x_0-1} x_0\ne 1}$

${\displaystyle f\left(x_0\right)=\frac{x_0^3-1}{x_0-1} x_0\ne 1}$

∴ ${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)=f\left(x_0\right)}$

f(x) is continuous at x = x₀.

Since x₀ is arbitrary f(x) is continuous at all points of ${\displaystyle (1,\infty )}$.

Case (iii) Let x₀ = 1 then

${\displaystyle \underset{x\to 1}{lim}f\left(x\right)= \underset{x\to 1}{lim}2}$ = 2

${\displaystyle f\left(1\right)}$ = 2

∴ ${\displaystyle \underset{x\to 1}{lim}f\left(x\right)=f\text{'}\left(1\right)}$

Hence, f(x) is continuous at x = 1.

Using all the three cases, we have f(x) is continuous at all the points of R.