# 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
$\displaystyle \begin{array}{l} y = 4^x \\ \frac{{dy}}{{dx}} = k.4^x \\ \end{array}$

This is my attempt

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

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

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
$\displaystyle \begin{array}{l} y = 4^x \\ \frac{{dy}}{{dx}} = k.4^x \\ \end{array}$

This is my attempt

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

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

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

Can you see a simplification ?

3. Yes sorry they should be negative signs

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