# How do you compute the instantaneous rate of change of f(x)=sqrt(4x+1) at the point (2,3) as a limit of the average rate of change of f(x) on the interval [2,2+h]?

##### 1 Answer
Mar 27, 2015

Thank you for being specific about the method you want!

As you probably know, you need to find:

${\lim}_{h \rightarrow 0} \frac{f \left(2 + h\right) - f \left(2\right)}{h} = {\lim}_{h \rightarrow 0} \frac{\sqrt{4 \left(2 + h\right) + 1} - \sqrt{4 \left(2\right) + 1}}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\sqrt{4 h + 9} - 3}{h}$ (and here's where the question arises)

An attempt to evaluate the limit by substitution results in the indeterminate form $\frac{0}{0}$.

Learn this trick (technique) try rationalizing the numerator. (Sometimes you have to try something and just see if it helps. That's how mathematics works beyond the elementary levels.)

$\frac{\sqrt{4 h + 9} - 3}{h} = \frac{\left(\sqrt{4 h + 9} - 3\right)}{h} \frac{\left(\sqrt{4 h + 9} + 3\right)}{\left(\sqrt{4 h + 9} + 3\right)}$

=((4h+9)-9)/(h(sqrt(4h+9)+3))=(4h)/(h(sqrt(4h+9)+3) which is still indeterminate as $h \rightarrow 0$

For $h \ne 0$, (sqrt(4h+9)-3)/h=4/(sqrt(4h+9)+3 So we can evaluate the limit.

${\lim}_{h \rightarrow 0} \frac{\sqrt{4 h + 9} - 3}{h} = {\lim}_{h \rightarrow 0} \frac{4}{\sqrt{4 h + 9} + 3} = \frac{4}{3 + 3} = \frac{2}{3}$

.

Without all the discussion, it looks like this:

${\lim}_{h \rightarrow 0} \frac{f \left(2 + h\right) - f \left(2\right)}{h} = {\lim}_{h \rightarrow 0} \frac{\sqrt{4 \left(2 + h\right) + 1} - \sqrt{4 \left(2\right) + 1}}{h}$

$= {\lim}_{h \rightarrow 0} \frac{\sqrt{4 h + 9} - 3}{h} = {\lim}_{h \rightarrow 0} \frac{\left(\sqrt{4 h + 9} - 3\right)}{h} \frac{\left(\sqrt{4 h + 9} + 3\right)}{\left(\sqrt{4 h + 9} + 3\right)}$

$= {\lim}_{h \rightarrow 0} \frac{\left(4 h + 9\right) - 9}{h \left(\sqrt{4 h + 9} + 3\right)} = {\lim}_{h \rightarrow 0} \frac{4}{\sqrt{4 h + 9} + 3} = \frac{4}{3 + 3} = \frac{2}{3}$
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