## Precalculus (6th Edition) Blitzer

Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}$. In making the table, choose the value of x close to 0 from the left and from the right as x approaches 0. As x approaches 0 from the left, arbitrarily start with $x=-1$. Then select two additional values of x that are closer to 0, but still less than 0; we choose $-0.1$ and $-0.01$. Evaluate f at each chosen value of x to obtain the corresponding values of $f\left( x \right)$. All the values of x are in radians. At $x=-1$, Substitute, $x=-1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}$. Therefore, $\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( -1 \right)}^{2}}}{\sec \left( -1 \right)-1}$ Apply the trigonometric property, $\sec \left( -\theta \right)=\sec \left( \theta \right)$. \begin{align} & \underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( -1 \right)}^{2}}}{\sec \left( 1 \right)-1} \\ & =\frac{1}{1.8508-1} \\ & =\frac{1}{0.8508} \\ & =1.1753 \end{align} Now, at $x=-0.1$ Substitute, $x=-0.1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}$. $\underset{x\to -0.1}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( -0.1 \right)}^{2}}}{\sec \left( -0.1 \right)-1}$ Apply the trigonometric property, $\sec \left( -\theta \right)=\sec \left( \theta \right)$ Therefore, \begin{align} & \underset{x\to -0.1}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( -0.1 \right)}^{2}}}{\sec \left( 0.1 \right)-1} \\ & =\frac{0.01}{1.0050-1} \\ & =\frac{0.01}{0.0050} \\ & =1.9917 \end{align} At $x=-0.01$ Substitute, $x=-0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}$. $\underset{x\to -0.01}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( -0.01 \right)}^{2}}}{\sec \left( -0.01 \right)-1}$ Apply the trigonometric property, $\sec \left( -\theta \right)=\sec \left( \theta \right)$ \begin{align} & \underset{x\to -0.01}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( -0.01 \right)}^{2}}}{\sec \left( 0.01 \right)-1} \\ & =\frac{0.0001}{1.00005-1} \\ & =\frac{0.01}{0.00005} \\ & =1.9999 \end{align} Now, as x approaches $0$ from the right, arbitrarily start with $x=1$. Then select two additional values of x that are closer to $0$, but still greater than $0$; we choose $0.1$ and $0.01$. At $x=1$ Substitute, $x=1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}$. Therefore, \begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( 1 \right)}^{2}}}{\sec \left( 1 \right)-1} \\ & =\frac{1}{1.8508-1} \\ & =\frac{1}{0.8508} \\ & =1.1753 \end{align} At $x=0.1$ Substitute, $x=0.1$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}$. Therefore, \begin{align} & \underset{x\to 0.1}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( 0.1 \right)}^{2}}}{\sec \left( 0.1 \right)-1} \\ & =\frac{0.01}{1.0050-1} \\ & =\frac{0.01}{0.0050} \\ & =1.9917 \end{align} At $x=0.01$ Substitute, $x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}$. \begin{align} & \underset{x\to 0.01}{\mathop{\lim }}\,\frac{{{x}^{2}}}{\sec x-1}=\frac{{{\left( 0.01 \right)}^{2}}}{\sec \left( 0.01 \right)-1} \\ & =\frac{0.0001}{1.00005-1} \\ & =\frac{0.01}{0.00005} \\ & =1.9999 \end{align} The limit of $\frac{{{x}^{2}}}{\sec x-1}$ as x approaches $0$ equals the number 2.