Precalculus (6th Edition) Blitzer

The value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{4g\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}$ is $4$.
To find the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{4g\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}$, find the value of $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ and the value of $\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)$ It is seen from the table that as the value of x nears $0$ from the left or right, the value of the function $f\left( x \right)$ nears $-1$. Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-1$ It is seen from the table that as the value of x nears $0$ from the left or right, the value of the function $g\left( x \right)$ nears $1$. Thus, $\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)=1$ Now find the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{4g\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}$, \begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\frac{4g\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}=\frac{4\underset{x\to 0}{\mathop{\lim }}\,g\left( x \right)}{{{\left[ \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right) \right]}^{2}}} \\ & =\frac{4\left( 1 \right)}{{{\left[ -1 \right]}^{2}}} \\ & =\frac{4}{1} \\ & =4 \end{align} Thus, the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{4g\left( x \right)}{{{\left[ f\left( x \right) \right]}^{2}}}$ is $4$.