## Precalculus (6th Edition) Blitzer

The complete statement is “$\underset{x\to a}{\mathop{\lim }}\,\text{ }\frac{f\left( x \right)}{g\left( x \right)}=$$\frac{L}{M}$$ M\ne 0$.
The limit of the quotient of two functions equals the quotient of their limits. Also, 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 $\underset{x\to a}{\mathop{\lim }}\,\text{ }\frac{f\left( x \right)}{g\left( x \right)}=\frac{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)}=\frac{L}{M}\text{ };M\ne 0$ 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. For example: Let $f\left( x \right)=x$ and $g\left( x \right)=2$, \begin{align} & \underset{x\to 7}{\mathop{\lim }}\,\frac{f\left( x \right)}{g\left( x \right)}=\frac{\underset{x\to 7}{\mathop{\lim }}\,f\left( x \right)}{\underset{x\to 7}{\mathop{\lim }}\,g\left( x \right)} \\ & =\frac{\underset{x\to 7}{\mathop{\lim }}\,x}{\underset{x\to 7}{\mathop{\lim }}\,2} \\ & =\frac{7}{2} \end{align} Therefore, the complete fill for the blank in the statement “$\underset{x\to a}{\mathop{\lim }}\,\text{ }\frac{f\left( x \right)}{g\left( x \right)}=\frac{L}{M}$,$M\ne 0$”.