Answer
The value of $\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}$ is $\frac{5}{4}$.
Work Step by Step
Consider the expression:
$\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}$,
To find the limit of the function, substitute $-3$ for x in above function,
The function becomes,
$\begin{align}
& \underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}=\frac{{{\left( -3 \right)}^{2}}+\left( -3 \right)-6}{{{\left( -3 \right)}^{2}}+2\left( -3 \right)-3} \\
& =\frac{9-3-6}{9-6-3} \\
& =\frac{0}{0}
\end{align}$
Here, the value is not defined at
$x=-3$
Therefore, direct substitution is not possible.
Therefore,
$\begin{align}
& \underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}=\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+3x-2x-6}{{{x}^{2}}+3x-x-3} \\
& =\underset{x\to -3}{\mathop{\lim }}\,\frac{\left( {{x}^{2}}+3x \right)+\left( -2x-6 \right)}{\left( {{x}^{2}}+3x \right)+\left( -x-3 \right)} \\
& =\underset{x\to -3}{\mathop{\lim }}\,\frac{x\left( x+3 \right)-2\left( x+3 \right)}{x\left( x+3 \right)-\left( x+3 \right)} \\
& =\underset{x\to -3}{\mathop{\lim }}\,\frac{\left( x+3 \right)\left( x-2 \right)}{\left( x+3 \right)\left( x-1 \right)}
\end{align}$
Eliminate the common terms,
$\underset{x\to -3}{\mathop{\lim }}\,\frac{\left( x+3 \right)\left( x-2 \right)}{\left( x+3 \right)\left( x-1 \right)}=\underset{x\to -3}{\mathop{\lim }}\,\frac{x-2}{x-1}$
Now substitute $x=-3$ in the above function as shown below,
$\begin{align}
& \underset{x\to -3}{\mathop{\lim }}\,\frac{x-2}{x-1}=\frac{-3-2}{-3-1} \\
& =\frac{-5}{-4} \\
& =\frac{5}{4}
\end{align}$
Therefore, the value of $\underset{x\to -3}{\mathop{\lim }}\,\frac{{{x}^{2}}+x-6}{{{x}^{2}}+2x-3}$ is $\frac{5}{4}$.
