## Precalculus (6th Edition) Blitzer

Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}$. 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 $x=-0.01$. Then select two additional values of x that are closer to 0, but still less than 0; we choose $-0.001$ and $-0.0001$. 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 $x=-0.01$, Substitute, $x=-0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}$. Therefore, \begin{align} & \underset{x\to -0.01}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}=\frac{\sin {{\left( -0.01 \right)}^{2}}}{\left( -0.01 \right)} \\ & =\frac{\sin \left( 1\times {{10}^{-4}} \right)}{\left( -0.01 \right)} \\ & =\frac{0.0001}{-0.01} \\ & =-0.01 \end{align} Now, at $x=-0.001$ Substitute, $x=-0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}$. \begin{align} & \underset{x\to -0.001}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}=\frac{\sin {{\left( -0.001 \right)}^{2}}}{\left( -0.001 \right)} \\ & =\frac{\sin \left( 1\times {{10}^{-6}} \right)}{\left( -0.001 \right)} \\ & =-\frac{0.000001}{0.001} \\ & =-0.001 \end{align} At $x=-0.0001$ Substitute, $x=-0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}$. Therefore, \begin{align} & \underset{x\to -0.0001}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}=\frac{\sin {{\left( -0.0001 \right)}^{2}}}{\left( -0.0001 \right)} \\ & =\frac{\sin \left( 1\times {{10}^{-8}} \right)}{\left( -0.0001 \right)} \\ & =-\frac{0.00000001}{0.0001} \\ & =-0.0001 \end{align} Now, as x approaches $0$ from the right, arbitrarily start with $x=0.01$. Then select two additional values of x that are closer to $0$, but still greater than $0$; we choose $0.001$ and $0.0001$. At $x=0.01$ Substitute, $x=0.01$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}$. Therefore, \begin{align} & \underset{x\to 0.01}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}=\frac{\sin {{\left( 0.01 \right)}^{2}}}{\left( 0.01 \right)} \\ & =\frac{\sin \left( 1\times {{10}^{-4}} \right)}{\left( 0.01 \right)} \\ & =\frac{0.0001}{0.01} \\ & =0.01 \end{align} Now, at $x=0.001$ Substitute, $x=0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}$. Therefore, \begin{align} & \underset{x\to 0.001}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}=\frac{\sin {{\left( 0.001 \right)}^{2}}}{\left( 0.001 \right)} \\ & =\frac{\sin \left( 1\times {{10}^{-6}} \right)}{\left( 0.001 \right)} \\ & =\frac{0.000001}{0.001} \\ & =0.001 \end{align} At $x=0.0001$ Substitute, $x=0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}$. \begin{align} & \underset{x\to 0.0001}{\mathop{\lim }}\,\frac{\sin {{x}^{2}}}{x}=\frac{\sin {{\left( 0.0001 \right)}^{2}}}{\left( 0.0001 \right)} \\ & =\frac{\sin \left( 1\times {{10}^{-8}} \right)}{\left( 0.0001 \right)} \\ & =\frac{0.00000001}{0.0001} \\ & =0.0001 \end{align} The limit of $\frac{\sin {{x}^{2}}}{x}$ as x approaches $0$ equals the number 0.