## Precalculus (6th Edition) Blitzer

Consider the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$ . In making the table, choose the value of x close to 2 from the left and from the right as x approaches 2. As x approaches 2 from the left, arbitrarily start with $x=1.99$. Then select two additional values of x that are closer to 2, but still less than 2; we choose 1.999 and 1.9999. Evaluate f at each chosen value of x to obtain the corresponding values of $f\left( x \right)$. At $x=1.99$, Substitute, $x=1.99$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$. Therefore, \begin{align} & \underset{x\to 1.99}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 1.99 \right)}^{2}} \\ & =5\left( 3.9601 \right) \\ & =19.801 \end{align} Now, at $x=1.999$ Substitute, $x=1.999$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$. \begin{align} & \underset{x\to 1.999}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 1.999 \right)}^{2}} \\ & =5\left( 3.9960 \right) \\ & =19.980 \end{align} At $x=1.9999$ Substitute, $x=1.9999$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$. \begin{align} & \underset{x\to 1.9999}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 1.9999 \right)}^{2}} \\ & =5\left( 3.9996 \right) \\ & =19.998 \end{align} Now, as x approaches 2 from the right, arbitrarily start with $x=2.01$. Then select two additional values of x that are closer to 2, but still greater than 2; we choose 2.001 and 2.0001. At $x=2.01$ Substitute, $x=2.01$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$ \begin{align} & \underset{x\to 2.01}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 2.01 \right)}^{2}} \\ & =5\left( 4.0401 \right) \\ & =20.201 \end{align} Now, at $x=2.001$ Substitute, $x=2.001$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$ Therefore, \begin{align} & \underset{x\to 2.001}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 2.001 \right)}^{2}} \\ & =5\left( 4.0040 \right) \\ & =20.020 \end{align} At $x=2.0001$ Substitute, $x=2.0001$ in the provided limit notation, $\underset{x\to 2}{\mathop{\lim }}\,5{{x}^{2}}$ \begin{align} & \underset{x\to 2.0001}{\mathop{\lim }}\,5{{x}^{2}}=5{{\left( 2.0001 \right)}^{2}} \\ & =5\left( 4.0004 \right) \\ & =20.002 \end{align}