**What kind of discontinuity does f(x)= { 2, -5 < x < 0 cos(x) +1, 0 < x < 5 have at x=0**

The given piecewise function is:

\[
f(x) = 
\begin{cases}
2 & \text{if } -5 < x < 0 \\
\cos(x) + 1 & \text{if } 0 < x < 5
\end{cases}
\]

We are asked to determine the type of discontinuity at \(x = 0\). Let us proceed step-by-step.

### 1. Analyze the Left-Hand and Right-Hand Limits:
- **Left-hand limit** as \(x \to 0^{-}\) (approaching 0 from the left):
  \[
  \lim_{x \to 0^{-}} f(x) = 2
  \]
  Since \(f(x) = 2\) for all \(x\) in \((-5, 0)\), the limit from the left is 2.

- **Right-hand limit** as \(x \to 0^{+}\) (approaching 0 from the right):
  \[
  \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} (\cos(x) + 1) = \cos(0) + 1 = 2
  \]

### 2. Check the Function Value at \(x = 0\):
The function is **not defined** at \(x = 0\) because the piecewise function only covers the intervals \( -5 < x < 0 \) and \( 0 < x < 5 \). Hence:

\[
f(0) \text{ is undefined.}
\]

### 3. Conclusion:
- Both the left-hand limit and the right-hand limit exist and are equal:
  \[
  \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = 2
  \]
- However, the function is not defined at \(x = 0\).

Thus, the function has a **removable discontinuity** at \(x = 0\). A removable discontinuity occurs when the limit exists at a point but the function is either not defined or defined incorrectly at that point.

---

Do you need further clarification or details?  
Here are five related questions you might explore:

1. What is the difference between removable and jump discontinuities?
2. How can we "remove" a removable discontinuity?
3. What happens if the left-hand and right-hand limits are unequal at a point?
4. What type of discontinuity does \(\tan(x)\) have at \(x = \frac{\pi}{2}\)?
5. How do continuous and differentiable functions relate?

**Tip:** A function with a removable discontinuity can be made continuous by redefining the value at the discontinuous point.

What kind of discontinuity does f(x)= { 2, -5 < x < 0 cos(x) +1, 0 < x < 5 have at x=0?

Explore how to identify the type of discontinuity at x=0 for the piecewise function f(x) = { 2 for -5 < x < 0, cos(x) + 1 for 0 < x < 5}. This detailed analysis uses limits and explores removable discontinuities, suitable for advanced high school or early college students.

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