How do you use the limit definition to find the slope of the tangent line to the graph f(x)=x+2 at x=2?

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Patricia Duffy · Accepted Answer

By definition, the slope of the tangent line to the graph of a given function f(x) at a given point $x_{0}$ is the derivative $f^{'} (x_{0})$ , which in turn is defined as follows:
$f^{'} (x_{0}) = lim_{h \to 0} \frac{f (x_{0} + h) - f (x_{0})}{h}$
In your case, $f (x) = \sqrt{x + 2}$ and $x_{0} = 2$ . The definition becomes
$f^{'} (2) = lim_{h \to 0} \frac{\sqrt{(2 + h) + 2} - \sqrt{2 + 2}}{h} = lim_{h \to 0} \frac{\sqrt{4 + h} - 2}{h}$
To solve this limit we may "razionalize" the expression:
$\frac{\sqrt{4 + h} - 2}{h} = \frac{\sqrt{4 + h} - 2}{h} \cdot \frac{\sqrt{4 + h} + 2}{\sqrt{4 + h} + 2} = \frac{(4 + h) - 4}{\sqrt{4 + h} + 2} = \frac{h}{\sqrt{4 + h} + 2}$
The limit becomes
$f^{'} (2) = lim_{h \to 0} \frac{h}{h (\sqrt{4 + h} + 2)} = lim_{h \to 0} \frac{1}{\sqrt{4 + h} + 2} = \frac{1}{\sqrt{4 + 0} + 2} = \frac{1}{2 + 2}$
$= \frac{1}{4}$

How do you use the limit definition to find the

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