## Precalculus (6th Edition) Blitzer

The value of $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$ for $f\left( x \right)={{x}^{2}}+2x-3$, $a=1$ is 4.
The function is given as $f\left( x \right)={{x}^{2}}+2x-3$ As $a=1$, \begin{align} & f\left( a+h \right)=f\left( 1+h \right) \\ & ={{\left( 1+h \right)}^{2}}+2\left( 1+h \right)-3 \\ & ={{1}^{2}}+\left( 2\times 1\times h \right)+{{h}^{2}}+2+2h-3 \\ & =1+2h+{{h}^{2}}+2+2h-3 \end{align} Solve further: $f\left( a+h \right)={{h}^{2}}+4h$ \begin{align} & f\left( a \right)=f\left( 1 \right) \\ & ={{1}^{2}}+\left( 2\times 1 \right)-3 \\ & =1+2-3 \\ & =0 \end{align} Consider the provided limit $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$. Substitute the values of $f\left( a+h \right)$ and $f\left( a \right)$ \begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}+4h-0}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}+4h}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( h+4 \right)}{h} \end{align} Cancel out the common factor $h$ from both the numerator and denominator. \begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\left( h+4 \right)}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( h+4 \right) \end{align} Using limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ \begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\left( h+4 \right) \\ & =\underset{h\to 0}{\mathop{\lim }}\,h+\underset{h\to 0}{\mathop{\lim }}\,4 \end{align} Use the property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{ }$ in $\underset{h\to 0}{\mathop{\lim }}\,4$: \begin{align} & \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}=0+4 \\ & =4 \end{align} Thus, the value of $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$ for $f\left( x \right)={{x}^{2}}+2x-3$, $a=1$ is 4.