Question

Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
lim √9 + h - 3 / h
h-->0

Answer 1

Answer:

a = 0

f(h) = $\frac{\sqrt{9+h} - 3}{h}$

limit of the function is 1/6

Step-by-step explanation:

The general form representing limit of a function is expressed as shown below;

$\lim_{h \to a} f(h)$ where a is the value that h will take and use in the function f(h). It can be expressed in words as limit of function f as h tends to a. Comparing the genaral form of the limit to the limit given in question $\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}$, it can be seen that a = 0 and f(h) = $\frac{\sqrt{9+h} - 3}{h}$

Taking the limit of the function

$\lim_{h \to 0} \frac{\sqrt{9+h} -3}{h}\\= \frac{\sqrt{9+0}-3 }{0}\\= \frac{0}{0}(indeterminate)$

Applying l'hopital rule

$\lim_{h \to 0} \frac{\frac{d}{dh} (\sqrt{9+h} - 3)} {\frac{d}{dh} (h)}\\= \lim_{h \to 0} \frac{1}{2} (9+h)^{-1/2} /1\\=\frac{1}{2} (9+0)^{-1/2}\\= \frac{1}{2} * \frac{1}{\sqrt{9} } \\= 1/2 * 1/3\\= 1/6$