# Thread: find k in form of limit

1. ## find k in form of limit

Question asks to find k in the form of a limit of
$
\begin{array}{l}
y = 4^x \\
\frac{{dy}}{{dx}} = k.4^x \\
\end{array}
$

This is my attempt

$
\begin{array}{l}
\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{4^{a + h} + 4^a }}{h} \\
k.4^a = \frac{{4^{a + h} + 4^a }}{h} \\
k = \frac{1}{{4^a }}\frac{{4^{a + h} + 4^a }}{h} \\
\end{array}
$

$
\mathop {\lim }\limits_{h \to 0} \frac{{4^h + 1}}{h}
$

I'm not sure how to get to this.

2. Hello,
Originally Posted by Craka
Question asks to find k in the form of a limit of
$
\begin{array}{l}
y = 4^x \\
\frac{{dy}}{{dx}} = k.4^x \\
\end{array}
$

This is my attempt

$
\begin{array}{l}
\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{4^{a + h} {\color{red}-} 4^a }}{h} \\
k.4^a = \frac{{4^{a + h} {\color{red}-} 4^a }}{h} \\
k = \frac{1}{{4^a }}\frac{{4^{a + h} {\color{red}-} 4^a }}{h} \\
\end{array}
$

$
\mathop {\lim }\limits_{h \to 0} \frac{{4^h {\color{red}-} 1}}{h}
$

I'm not sure how to get to this.
I guess these are - signs and not + (see the red parts). And you are very near to the solution !

Note that $4^{a+h}-4^a=4^a \cdot 4^h-4^a=4^a \cdot \left(4^h-1\right)$

Can you see a simplification ?

3. Yes sorry they should be negative signs

$
\begin{array}{l}
\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{4^{a + h} - 4^a }}{h} \\
k.4^a = \frac{{4^{a + h} - 4^a }}{h} \\
k = \frac{1}{{4^a }}\frac{{4^{a + h} - 4^a }}{h} \\
\end{array}
$