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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity3 Marks
Question
Show that the following functions have a continuous extension to the point where $f(x)$ is not defined. Also, find the extension:$f(x)=\frac{x^2-1}{x^3+1}$, for $x \neq-1$
✓

Answer

$f(x)=\frac{x^2-1}{x^3+1} \text {, for } x \neq-1$
Here, $f(-1)$ is not defined.
Consider,
$\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}\left(\frac{x^2-1}{x^3+1}\right)$
$=\lim _{x \rightarrow-1} \frac{(x+1)(x-1)}{(x+1)\left(x^2-x+1\right)}$
$=\lim _{x \rightarrow-1} \frac{x-1}{x^2-x+1} \cdots\left[\begin{array}{l} \because x \rightarrow-1, x \neq-1 \\
\therefore x+1 \neq 0\end{array}\right]$
$=\frac{-1-1}{(-1)^2-(-1)+1}=-\frac{2}{3}$
$=\lim _{x \rightarrow-1} \frac{x-1}{x^2-x+1} \cdots\left[\begin{array}{l} \because x \rightarrow-1, x \neq-1 \\ \therefore x+1 \neq 0\end{array}\right]$
$=\frac{-1-1}{(-1)^2-(-1)+1}=-\frac{2}{3}$
But $f(-1)$ is not defined.
$\therefore f ( x )$ has a removable discontinuity at $x =-1$.
$\therefore$ The extension of the original function is
$f(x)=\frac{x^2-1}{x^3+1}, x \neq-1$
$=-\frac{2}{3}, x=-1$
$\therefore f(x)$ is continuous at $x=-\frac{2}{3}$
$\therefore f ( x )$ is continuous at $x =-\frac{2}{3}$

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