Precalculus (6th Edition) Blitzer

The limit $\underset{x\to 2}{\mathop{\lim }}\,{{\left( 3{{x}^{2}}-10 \right)}^{3}}$ is 8. And the corresponding limit property is $\underset{x\to a}{\mathop{\lim }}\,{{\left[ f\left( x \right) \right]}^{n}}={{\left[ \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) \right]}^{n}}={{L}^{n}}$, where $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$ and $n$ is any integer.
Consider the given limit, $\underset{x\to 2}{\mathop{\lim }}\,{{\left( 3{{x}^{2}}-10 \right)}^{3}}$. First, find $\underset{x\to 2}{\mathop{\lim }}\,\left( 3{{x}^{2}}-10 \right)$. \begin{align} & \underset{x\to 2}{\mathop{\lim }}\,\left( 3{{x}^{2}}-10 \right)=\left( 3\times {{2}^{2}} \right)-10 \\ & =3\times 4-10 \\ & =12-10 \\ & =2 \end{align} The limit that is required is calculated by taking this limit, 2, and raising it to the third power. Thus, $\underset{x\to 2}{\mathop{\lim }}\,{{\left( 3{{x}^{2}}-10 \right)}^{3}}={{\left[ \underset{x\to 2}{\mathop{\lim }}\,\left( 3{{x}^{2}}-10 \right) \right]}^{3}}={{2}^{3}}=8$. In limit notation, the corresponding limit property is If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$ and $n$ is any integer, then $\underset{x\to a}{\mathop{\lim }}\,{{\left[ f\left( x \right) \right]}^{n}}={{\left[ \underset{x\to a}{\mathop{\lim }}\,f\left( x \right) \right]}^{n}}={{L}^{n}}$ In words, the limit of a function to a power is found by taking the limit of the function and then raising this limit to the power.