The expression in the image you provided is the derivative of a function $f(x)$ at a point $a$ using the definition of the derivative:

$f'(a) = \lim_{h \to 0} \frac{\frac{2}{1+h} - 2}{h}$

This appears to represent the derivative of the function $f(x) = \frac{2}{x}$, evaluated at $a = 1$, based on the structure of the difference quotient. Let's verify this and then find the equation of the tangent line.

Step-by-step Solution:

1. Find the derivative $f'(1)$

We are given the limit expression:

$f'(1) = \lim_{h \to 0} \frac{\frac{2}{1+h} - 2}{h}$

Simplify the numerator: $\frac{2}{1+h} - 2 = \frac{2 - 2(1+h)}{1+h} = \frac{2 - 2 - 2h}{1+h} = \frac{-2h}{1+h}$

Now, substitute this into the limit expression: $f'(1) = \lim_{h \to 0} \frac{\frac{-2h}{1+h}}{h}$

Simplify further: $f'(1) = \lim_{h \to 0} \frac{-2h}{h(1+h)} = \lim_{h \to 0} \frac{-2}{1+h}$

As $h \to 0$, this becomes: $f'(1) = \frac{-2}{1+0} = -2$

2. Equation of the tangent line at $x = 1$

The equation of the tangent line at a point $(a, f(a))$ is given by the point-slope form: $y - f(a) = f'(a)(x - a)$

We know that $f(x) = \frac{2}{x}$, so: $f(1) = \frac{2}{1} = 2$ And we already found that $f'(1) = -2$. Therefore, the equation of the tangent line is: $y - 2 = -2(x - 1)$ Simplifying: $y - 2 = -2x + 2$ $y = -2x + 4$

Conclusion:

The equation of the tangent line to the curve $f(x) = \frac{2}{x}$ at $x = 1$ is: $y = -2x + 4$

Would you like more details on any part of the process?

Follow-up Questions:

1. What is the geometric meaning of a derivative at a point?
2. How would the tangent line change for other values of $a$?
3. Can you find the second derivative of $f(x) = \frac{2}{x}$?
4. What does the derivative of $f(x)$ tell you about the concavity of the function?
5. How does the point-slope form of a line relate to the derivative of a function?

Tip:

When simplifying limits, always check if you can cancel common terms before directly evaluating the limit.