# Limit Definition of Derivative

## Key Questions

• Limit Definition of $f ' \left(a\right)$

$f ' \left(a\right) = {\lim}_{h \to 0} \frac{f \left(a + h\right) - f \left(a\right)}{h}$

or

$f ' \left(a\right) = {\lim}_{x \to a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$

• $f \left(x\right) = c$ is a constant function, so its value stays the same regardless of the x-value. In particular, $f \left(x + h\right) = c$.

By the definition of the derivative,

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$= {\lim}_{h \to 0} \frac{c - c}{h}$

$= {\lim}_{h \to 0} 0$

$= 0$

• Remember that the limit definition of the derivative goes like this:
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$.
So, for the posted function, we have
$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{m \left(x + h\right) + b - \left[m x + b\right]}{h}$
By multiplying out the numerator,
$= {\lim}_{h \rightarrow 0} \frac{m x + m h + b - m x - b}{h}$
By cancelling out $m x$'s and $b$'s,
$= {\lim}_{h \rightarrow 0} \frac{m h}{h}$
By cancellng out $h$'s,
$= {\lim}_{h \rightarrow 0} m = m$
Hence, $f ' \left(x\right) = m$.

The answer above makes sense since the derivative tells us about the slope of the tangent line to the graph of $f$, and the slope of the linear function (its graph is a line) is $m$.

• Yes, there is a difference since the first limit is defined at $x = 0$, but the second one is not.

I hope that this was helpful.