To find the derivative of \( f(x) = \frac{1}{x} + 2 \) at \( x = 2 \) using the primary definition of a derivative, we start with the definition:
\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
In this case, \( a = 2 \). First, we calculate \( f(2) \):
\[ f(2) = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2} \]
Now we need to find \( f(2+h) \):
\[ f(2+h) = \frac{1}{2+h} + 2 \]
Substituting both \( f(2+h) \) and \( f(2) \) into the derivative definition gives:
\[ f'(2) = \lim_{h \to 0} \frac{\left(\frac{1}{2+h} + 2\right) - \frac{5}{2}}{h} \]
Simplifying the above expression:
\[ = \lim_{h \to 0} \frac{\frac{1}{2+h} + 2 - \frac{5}{2}}{h} \]
Now, combine the constant terms:
\[ 2 = \frac{4}{2} \implies 2 - \frac{5}{2} = \frac{4}{2} - \frac{5}{2} = \frac{-1}{2} \]
So the expression becomes:
\[ = \lim_{h \to 0} \frac{\frac{1}{2+h} - \frac{1}{2}}{h} \]
Now simplify the fraction:
\[ = \lim_{h \to 0} \frac{\frac{1 - (2+h)}{(2+h) \cdot 2}}{h} \]
This simplifies to:
\[ = \lim_{h \to 0} \frac{\frac{-h}{(2+h) \cdot 2}}{h} \]
Cancel \( h \) in the numerator and denominator (noting that this is valid as long as \( h \neq 0 \)):
\[ = \lim_{h \to 0} \frac{-1}{(2+h)\cdot 2} \]
Now, taking the limit as \( h \) approaches 0:
\[ = \frac{-1}{(2+0)\cdot 2} = \frac{-1}{2 \cdot 2} = \frac{-1}{4} \]
Therefore, the derivative of \( f(x) = \frac{1}{x} + 2 \) at \( x = 2 \) is:
\[ f'(2) = -\frac{1}{4} \]