## Precalculus (6th Edition) Blitzer

Consider the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$. 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, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$. Therefore, $\underset{x\to -0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( -0.01 \right)}^{2}}+\left( -0.01 \right)}{\sin \left( -0.01 \right)}$ Apply the trigonometric property, $\sin \left( -\theta \right)=-\sin \left( \theta \right)$ \begin{align} & \underset{x\to -0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 0.0001 \right)-0.01}{-\sin \left( 0.01 \right)} \\ & =\frac{0.0002-0.01}{-0.0099} \\ & =\frac{-9.8\times {{10}^{-3}}}{-0.0099} \\ & =0.9800 \end{align} Now, at $x=-0.001$ Substitute, $x=-0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$. $\underset{x\to -0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( -0.001 \right)}^{2}}+\left( -0.001 \right)}{\sin \left( -0.001 \right)}$ Apply the trigonometric property, $\sin \left( -\theta \right)=-\sin \left( \theta \right)$ Therefore, \begin{align} & \underset{x\to -0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 1\times {{10}^{-6}} \right)-0.001}{-\sin \left( 0.001 \right)} \\ & =\frac{\left( 2\times {{10}^{-6}} \right)-0.001}{-\left( 9.99\times {{10}^{-4}} \right)} \\ & =\frac{-9.98\times {{10}^{-4}}}{-\left( 9.99\times {{10}^{-4}} \right)} \\ & =0.9980 \end{align} At $x=-0.0001$ Substitute, $x=-0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$. $\underset{x\to -0.0001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.0001 \right)}^{2}}+\left( -0.0001 \right)}{\sin \left( -0.0001 \right)}$ Apply the trigonometric property, $\sin \left( -\theta \right)=-\sin \left( \theta \right)$ \begin{align} & \underset{x\to -0.0001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 1\times {{10}^{-8}} \right)-0.0001}{-\sin \left( 0.0001 \right)} \\ & =\frac{\left( 2\times {{10}^{-8}} \right)-0.0001}{-\left( 9.99\times {{10}^{-5}} \right)} \\ & =\frac{-9.998\times {{10}^{-5}}}{-\left( 9.99\times {{10}^{-5}} \right)} \\ & =0.9998 \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, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$. Therefore, $\underset{x\to 0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.01 \right)}^{2}}+\left( 0.01 \right)}{\sin \left( 0.01 \right)}$ \begin{align} & \underset{x\to 0.01}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2\left( 0.0001 \right)+0.01}{\sin \left( 0.01 \right)} \\ & =\frac{0.0002+0.01}{0.0099} \\ & =\frac{0.0102}{0.0099} \\ & =1.0200 \end{align} Now, at $x=0.001$ Substitute, $x=0.001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$ \begin{align} & \underset{x\to 0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.001 \right)}^{2}}+\left( 0.001 \right)}{\sin \left( 0.001 \right)} \\ & =\frac{2\left( 1\times {{10}^{-6}} \right)+0.001}{9.99\times {{10}^{-4}}} \\ & =\frac{\left( 2\times {{10}^{-6}} \right)+0.001}{9.99\times {{10}^{-4}}} \\ & =\frac{1.002\times {{10}^{-3}}}{9.99\times {{10}^{-4}}} \end{align} Further solve, $\underset{x\to 0.001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=1.0020$ At $x=0.0001$ Substitute, $x=0.0001$ in the provided limit notation, $\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}$. Therefore, \begin{align} & \underset{x\to 0.0001}{\mathop{\lim }}\,\frac{2{{x}^{2}}+x}{\sin x}=\frac{2{{\left( 0.0001 \right)}^{2}}+0.0001}{\sin \left( 0.0001 \right)} \\ & =\frac{2\left( 1\times {{10}^{-8}} \right)+0.0001}{9.99\times {{10}^{-5}}} \\ & =\frac{\left( 2\times {{10}^{-8}} \right)+0.0001}{\left( 9.99\times {{10}^{-5}} \right)} \\ & =\frac{1.0002\times {{10}^{-4}}}{\left( 9.99\times {{10}^{-5}} \right)} \\ & =1.0002 \end{align} The limit of $\frac{2{{x}^{2}}+x}{\sin x}$ as x approaches $0$ equals the number $1$.