## Precalculus (6th Edition) Blitzer

The equation “$\underset{x\to 5}{\mathop{\lim }}\,\text{ }\frac{{{x}^{2}}-25}{x-5}=\frac{\underset{x\to 5}{\mathop{\lim }}\,\text{ }\left( {{x}^{2}}-25 \right)}{\underset{x\to 5}{\mathop{\lim }}\,\text{ }\left( x-5 \right)}$” is false.
The limit of the quotient of two functions equals the quotient of their limits when the limit of the denominator is not zero. That is: The limit of a quotient: If $\underset{x\to a}{\mathop{\lim }}\,\text{ }f\left( x \right)=L\text{ and }\underset{x\to a}{\mathop{\lim }}\,\text{ }g\left( x \right)=M\text{ };M\ne 0$, then $\underset{x\to a}{\mathop{\lim }}\,\text{ }\frac{f\left( x \right)}{g\left( x \right)}=\frac{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)}=\frac{L}{M}\text{ };M\ne 0$ This is the quotient property. Before applying the quotient property, find the limit of the denominator. If this limit is not zero, apply the quotient property. And if the limit of the denominator is zero, the quotient property cannot be used. Now, the two functions in this limit problem are $f\left( x \right)={{x}^{2}}-25$ and $g\left( x \right)=x-5$. The limit of the quotient of these functions is $5$. The first step is to find the limit of the denominator, $g\left( x \right)$. \begin{align} & \underset{x\to 5}{\mathop{\lim }}\,\text{ }x-5=5-5 \\ & =0 \end{align} Since, the limit of the denominator is zero, the quotient property cannot be used. Hence, $\underset{x\to 5}{\mathop{\lim }}\,\text{ }\frac{{{x}^{2}}-25}{x-5}$ cannot be written as $\frac{\underset{x\to 5}{\mathop{\lim }}\,\text{ }\left( {{x}^{2}}-25 \right)}{\underset{x\to 5}{\mathop{\lim }}\,\text{ }\left( x-5 \right)}$. Therefore, the statement “$\underset{x\to 5}{\mathop{\lim }}\,\text{ }\frac{{{x}^{2}}-25}{x-5}=\frac{\underset{x\to 5}{\mathop{\lim }}\,\text{ }\left( {{x}^{2}}-25 \right)}{\underset{x\to 5}{\mathop{\lim }}\,\text{ }\left( x-5 \right)}$” is false.