hi why doesx+2arctan(1/x) become 0 when i put in x=0, but with limit it becomes pi instead?

The expression \( x + 2 \arctan\left(\frac{1}{x}\right) \) does indeed behave differently when evaluated directly at \( x = 0 \) versus using a limit as \( x \) approaches \( 0 \).

1. **Direct Substitution**: If you try to directly substitute \( x = 0 \) into \( x + 2 \arctan\left(\frac{1}{x}\right) \), you run into a problem. The term \( \arctan\left(\frac{1}{x}\right) \) becomes undefined because \( \frac{1}{x} \) is undefined at \( x = 0 \). So, we cannot directly evaluate this expression by plugging in \( x = 0 \).

2. **Using the Limit**: To understand the behavior as \( x \) approaches \( 0 \), we evaluate
   \[
   \lim_{x \to 0} \left( x + 2 \arctan\left(\frac{1}{x}\right) \right).
   \]
   
   Let's analyze this limit by considering the behavior of \( \arctan\left(\frac{1}{x}\right) \) as \( x \) approaches \( 0 \) from the left (\( x \to 0^- \)) and from the right (\( x \to 0^+ \)):

   - **As \( x \to 0^+ \)**, \( \frac{1}{x} \to +\infty \), so \( \arctan\left(\frac{1}{x}\right) \to \frac{\pi}{2} \).
   - **As \( x \to 0^- \)**, \( \frac{1}{x} \to -\infty \), so \( \arctan\left(\frac{1}{x}\right) \to -\frac{\pi}{2} \).

   Using these results, we can examine the behavior of the whole expression as \( x \to 0^+ \) and \( x \to 0^- \):

   - **For \( x \to 0^+ \)**:
     \[
     x + 2 \arctan\left(\frac{1}{x}\right) \approx 0 + 2 \cdot \frac{\pi}{2} = \pi.
     \]

   - **For \( x \to 0^- \)**:
     \[
     x + 2 \arctan\left(\frac{1}{x}\right) \approx 0 + 2 \cdot \left(-\frac{\pi}{2}\right) = -\pi.
     \]

   Since the limit depends on the direction from which \( x \) approaches \( 0 \), we conclude that
   \[
   \lim_{x \to 0} \left( x + 2 \arctan\left(\frac{1}{x}\right) \right)
   \]
   does not exist in the usual sense because it yields different results from the left and right (i.e., \( \pi \) from the right and \( -\pi \) from the left).

### Summary
The expression \( x + 2 \arctan\left(\frac{1}{x}\right) \) approaches different values as \( x \to 0 \) from the left and the right, leading to a **non-existent two-sided limit**.

Would you like a more detailed breakdown on one-sided limits, or have questions on limits for similar expressions?

1. How do one-sided limits work in general?
2. Why is the arctangent function involved in limits to infinity?
3. How do limits behave in piecewise functions?
4. Why does infinity cause discontinuities?
5. What are other ways to handle undefined expressions?

**Tip:** When dealing with expressions that contain terms like \( \frac{1}{x} \), examine limits from both sides to understand the behavior fully.

hi why does x + 2 arctan(1/x) become 0 when i put in x = 0, but with limit it becomes pi instead?

Explore why the expression x + 2 arctan(1/x) behaves differently when evaluated directly at x = 0 versus using a limit as x approaches 0. This explanation covers key limit concepts, trigonometric behavior at infinity, and the importance of one-sided limits. Ideal for students in Grades 11-12.

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