Answer
The statement “I can find the slope of the tangent line to the graph of $f\left( x \right)$ at $\left( 3,f\left( 3 \right) \right)$ using $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 3+h \right)-f\left( 3 \right)}{h}$ or finding $f'\left( x \right)$ and then replacing x with $3$” makes sense.
Work Step by Step
The slope of the tangent line to the graph of a function $f\left( x \right)$ 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.
Thus, the slope of the tangent line to the graph of a function $f\left( x \right)$ at $\left( 3,f\left( 3 \right) \right)$ is given by $\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 3+h \right)-f\left( 3 \right)}{h}$, as long as this limit exists.
The derivative of the function $f\left( x \right)$ is given by $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( a+h \right)-f\left( a \right)}{h}$ ,
Thus, the slope of the tangent line to the graph of $f\left( x \right)$ at $\left( 3,f\left( 3 \right) \right)$ is given by $f'\left( 3 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 3+h \right)-f\left( 3 \right)}{h}$; that is by finding $f'\left( x \right)$ and then replacing x with $3$.
Thus, the statement makes sense.
