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0/1 pt 5 19
question 7
given $f(x)=\frac{2}{x}$, find $f(x)$ using the limit definition of the derivative.
$f(x)=$

Explanation:

Step1: Recall limit definition of derivative

The limit definition of the derivative is:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step2: Substitute $f(x)=\frac{2}{x}$ into formula

First, find $f(x+h)=\frac{2}{x+h}$. Substitute into the expression:
$$f'(x) = \lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}$$

Step3: Combine numerator fractions

Find a common denominator $x(x+h)$ for the numerator:
$$\frac{2}{x+h} - \frac{2}{x} = \frac{2x - 2(x+h)}{x(x+h)} = \frac{2x - 2x - 2h}{x(x+h)} = \frac{-2h}{x(x+h)}$$
Now substitute back into the limit:
$$f'(x) = \lim_{h \to 0} \frac{\frac{-2h}{x(x+h)}}{h}$$

Step4: Simplify the complex fraction

Cancel $h$ (where $h
eq 0$):
$$f'(x) = \lim_{h \to 0} \frac{-2}{x(x+h)}$$

Step5: Evaluate the limit as $h \to 0$

Substitute $h=0$ into the expression:
$$f'(x) = \frac{-2}{x(x+0)} = \frac{-2}{x^2}$$

Answer:

$\boldsymbol{-\frac{2}{x^2}}$