evaluate the limit as x approaches 1 of tan^-1(x) - pi/4/tan(pix/4)-1 using l'hopitals rule

We are tasked with evaluating the limit:

\[
\lim_{{x \to 1}} \frac{\tan^{-1}(x) - \frac{\pi}{4}}{\tan\left(\frac{\pi x}{4}\right) - 1}
\]

We can apply L'Hôpital's Rule since this limit has the indeterminate form \( \frac{0}{0} \) as \( x \to 1 \).

### Step 1: Check if it's an indeterminate form
First, we check the values of the numerator and denominator at \( x = 1 \).

- For the numerator: \( \tan^{-1}(1) = \frac{\pi}{4} \), so
  \[
  \tan^{-1}(1) - \frac{\pi}{4} = 0.
  \]
  
- For the denominator: \( \tan\left(\frac{\pi}{4} \times 1\right) = \tan\left(\frac{\pi}{4}\right) = 1 \), so
  \[
  \tan\left(\frac{\pi}{4}\right) - 1 = 0.
  \]

Thus, we have the form \( \frac{0}{0} \), and L'Hôpital's Rule is applicable.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if a limit has the form \( \frac{0}{0} \), then we differentiate the numerator and denominator separately with respect to \( x \), and then take the limit of the new expression.

#### Numerator:
We need to differentiate \( \tan^{-1}(x) - \frac{\pi}{4} \). The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \), so the derivative of the numerator is:
\[
\frac{d}{dx} \left(\tan^{-1}(x) - \frac{\pi}{4}\right) = \frac{1}{1 + x^2}.
\]

#### Denominator:
We need to differentiate \( \tan\left(\frac{\pi x}{4}\right) - 1 \). The derivative of \( \tan\left(\frac{\pi x}{4}\right) \) is \( \frac{\pi}{4} \sec^2\left(\frac{\pi x}{4}\right) \), so the derivative of the denominator is:
\[
\frac{d}{dx} \left( \tan\left(\frac{\pi x}{4}\right) - 1 \right) = \frac{\pi}{4} \sec^2\left(\frac{\pi x}{4}\right).
\]

### Step 3: Compute the new limit
Now, we compute the new limit:
\[
\lim_{{x \to 1}} \frac{\frac{1}{1 + x^2}}{\frac{\pi}{4} \sec^2\left(\frac{\pi x}{4}\right)}.
\]

At \( x = 1 \), we substitute the values:

- The numerator: \( \frac{1}{1 + 1^2} = \frac{1}{2} \).
- The denominator: \( \sec^2\left(\frac{\pi}{4}\right) = 2 \) because \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \).

Thus, the limit becomes:
\[
\frac{\frac{1}{2}}{\frac{\pi}{4} \times 2} = \frac{\frac{1}{2}}{\frac{\pi}{2}} = \frac{1}{\pi}.
\]

### Final Answer:
\[
\lim_{{x \to 1}} \frac{\tan^{-1}(x) - \frac{\pi}{4}}{\tan\left(\frac{\pi x}{4}\right) - 1} = \frac{1}{\pi}.
\]

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Would you like more details on this solution, or do you have any questions? Here are a few related questions you might find interesting:

1. How does L'Hôpital's Rule apply to limits with other indeterminate forms like \( \frac{\infty}{\infty} \)?
2. Can you explain the behavior of the arctangent and tangent functions near their limits?
3. What are other techniques for evaluating limits involving trigonometric functions?
4. How do the derivatives of inverse trigonometric functions compare to standard trigonometric functions?
5. What happens to the value of a limit when the numerator and denominator both approach infinity?

**Tip:** Always check the form of the limit before deciding to use L'Hôpital’s Rule!

Learn how to evaluate the limit of tan^-1(x) - pi/4 / tan(pix/4) - 1 as x approaches 1 using L'Hopital's Rule. This problem involves applying differentiation to trigonometric and inverse trigonometric functions, offering an essential practice for advanced calculus students.

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