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The problem asks to show that the given piecewise function:

\[
f(x) = \begin{cases} 
\frac{x}{1+x} & \text{if } x \geq 0, \\
x^2 & \text{if } x < 0,
\end{cases}
\]
is not differentiable at \( x = 0 \) using the limit definition of the derivative.

### Step-by-Step Solution:

The derivative of \( f(x) \) at \( x = 0 \) is given by the limit definition of the derivative:
\[
f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h}.
\]

First, let's evaluate \( f(0) \) using the piecewise definition.

- For \( x \geq 0 \), \( f(x) = \frac{x}{1+x} \). So, \( f(0) = \frac{0}{1+0} = 0 \).
- For \( x < 0 \), \( f(x) = x^2 \). At \( x = 0 \), this also gives \( f(0) = 0^2 = 0 \).

Thus, \( f(0) = 0 \).

Now, compute the limit for \( h \to 0^+ \) (approaching 0 from the right):
\[
f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h \to 0^+} \frac{h}{h(1+h)} = \lim_{h \to 0^+} \frac{1}{1+h} = 1.
\]

Next, compute the limit for \( h \to 0^- \) (approaching 0 from the left):
\[
f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h^2 - 0}{h} = \lim_{h \to 0^-} h = 0.
\]

Since \( f'(0^+) = 1 \) and \( f'(0^-) = 0 \), the left-hand and right-hand limits are not equal. Therefore, \( f'(0) \) does not exist, meaning the function is **not differentiable at \( x = 0 \)**.

Would you like further details or explanations on any step? Here are some questions to consider:

1. What is the general concept behind the limit definition of a derivative?
2. Why do we need to evaluate limits from both sides to check differentiability at \( x = 0 \)?
3. How does the piecewise nature of the function affect the differentiability?
4. Could we say anything about the continuity of \( f(x) \) at \( x = 0 \)?
5. What are the geometric implications of the differing derivatives from the left and right?

**Tip:** Always check continuity first before checking differentiability at a point in piecewise functions.

Show that the function f(x) = { x/(1+x) if x ≥ 0, x^2 if x < 0 } is not differentiable at x = 0. Use the limit definition.

This page provides a detailed solution showing that the function f(x) = { x/(1+x) if x ≥ 0, x^2 if x < 0 } is not differentiable at x = 0 using the limit definition of the derivative. The solution evaluates the right-hand and left-hand limits and shows why the derivative does not exist at x = 0.

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