Chapter 11 - Section 11.1 - Finding Limits Using Tables and Graphs - Exercise Set - Page 1138: 14

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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.