# lim_(h->0)(25 xx 5^h-25)/h =  ?

Oct 22, 2017

$f \left(x\right) = {5}^{x}$
$f \left(a\right) = 25$
$\textcolor{w h i t e}{\text{----}} a = 2$

#### Explanation:

This is a limit definition of a derivative.

The formula for this is:

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

This tells us that

$f \left(a + h\right) = {5}^{2 + h}$

$f \left(a\right) = 25$

How can we figure out what $f \left(x\right)$ is? Well, in $f \left(a + h\right)$, we can see that the $h$ is in the exponent, so there's a good chance that the function is some sort of exponential function, probably with a base of 5 as well. So, let's try this:

Assume $f \left(x\right) = {5}^{x}$

This means that

$f \left(a + h\right) = {5}^{2 + h}$

$a + h = 2 + h$

$a = 2$

Now to check if this works out correctly.

$f \left(a\right) = f \left(2\right) = {5}^{2} = 25$, which we know is true from earlier.

Therefore, we can say that $f \left(x\right) = {5}^{x}$ and $f \left(a\right) = 25$.

Oct 22, 2017

See below.

#### Explanation:

Calling

$f \left(x\right) = {5}^{x}$ we have

${\lim}_{h \to 0} \frac{f \left(2 + h\right) - f \left(2\right)}{h} = f ' \left(2\right) = {\log}_{e} 5 \cdot {5}^{2} = 25 \cdot {\log}_{e} 5$

NOTE:

As we know $\frac{d}{\mathrm{dx}} {e}^{\alpha x} = \alpha {e}^{\alpha x}$ and

$a = {e}^{{\log}_{e} a}$ then ${a}^{x} = {e}^{{\log}_{e} a \cdot x}$ and then

$\frac{d}{\mathrm{dx}} {a}^{x} = \frac{d}{\mathrm{dx}} {e}^{{\log}_{e} a \cdot x} = {\log}_{e} a \cdot {e}^{{\log}_{e} a \cdot x} = {\log}_{e} a \cdot {a}^{x}$