## Precalculus (6th Edition) Blitzer

“The slope of the tangent line to the graph of a function $y=f\left( x \right)$ at $\left( a,f\left( a \right) \right)$ is given by $\underset{h\to 0}{\mathop{\lim }}\,$ $\frac{f\left( a+h \right)-f\left( a \right)}{h}$ provided that this limit exists. This limit also describes the instantaneous rate of change of the function f with respect to x at a.”
Consider the function $f\left( x \right)=2x$ and the point $\left( 1,2 \right)$ Here, $\left( a,f\left( a \right) \right)=\left( 1,2 \right)$ Now, compute the slope of the tangent line for the function $f\left( x \right)=2x$ as follows: Put $a=1$, ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$ To compute $f\left( 1+h \right)$, substitute $x=1+h$ in the function $f\left( x \right)=2x$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 2\left( 1+h \right) \right]-\left[ 2\left( 1 \right) \right]}{h}$ Now, simplify $2\left( 1+h \right)$ by using the distributive property $A\left( B+C \right)=AB+AC$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ 2+2h \right]-2}{h}$ Combine the like terms in the numerator. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{2h}{h}$ Cancel out h and apply the limits. \begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,2 \\ & =2 \end{align} Thus, the slope of the tangent to the graph of $f\left( x \right)=2x$ at $\left( 1,2 \right)$ is $2$. Hence, the slope ${{m}_{\tan }}$ is given by $\frac{f\left( a+h \right)-f\left( a \right)}{h}$ and it is the instantaneous rate the change of the function at $x=a$.