Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:

  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$. If a function is defined piecewise, continuity at the point where the definition changes requires the limit from the left to equal the limit from the right, and both to equal the function's value at that point.

• Limit of a Trigonometric Function:

  • $\lim_{h \to 0} \frac{\tan h}{h} = 1$.
  • $\tan(\frac{\pi}{4} + h) = \frac{\tan(\frac{\pi}{4}) + \tan h}{1 - \tan(\frac{\pi}{4}) \tan h} = \frac{1 + \tan h}{1 - \tan h}$.

• L'Hôpital's Rule (Optional, but useful for verification): If $\lim_{x \to c} \frac{g(x)}{h(x)}$ results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.

Step-by-Step Solution

Step 1: Understand the problem and the condition of continuity. The problem states that the function $f(x) = \frac{1 - \tan x}{4x - \pi}$ is continuous in the interval $\left[ {0,{\pi \over 2}} \right]$, and we need to find the value of $f\left( {{\pi \over 4}} \right)$. For $f(x)$ to be continuous at $x = \frac{\pi}{4}$, the value of the function at this point must be equal to the limit of the function as $x$ approaches $\frac{\pi}{4}$. That is, $f\left( {{\pi \over 4}} \right) = \lim_{x \to \frac{\pi}{4}} f(x)$.

Step 2: Evaluate the limit of the function as $x$ approaches $\frac{\pi}{4}$. We need to calculate $\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{4x - \pi}$. If we substitute $x = \frac{\pi}{4}$ directly into the expression, we get $\frac{1 - \tan(\frac{\pi}{4})}{4(\frac{\pi}{4}) - \pi} = \frac{1 - 1}{\pi - \pi} = \frac{0}{0}$, which is an indeterminate form. This indicates that we need to use limit evaluation techniques.

Step 3: Use a substitution to simplify the limit expression. Let $x = \frac{\pi}{4} + h$. As $x \to \frac{\pi}{4}$, $h \to 0$. We will evaluate the limit as $h \to 0$. Substitute $x = \frac{\pi}{4} + h$ into the function: $f\left( \frac{\pi}{4} + h \right) = \frac{1 - \tan\left(\frac{\pi}{4} + h\right)}{4\left(\frac{\pi}{4} + h\right) - \pi}$

Step 4: Simplify the numerator using the tangent addition formula. We know that $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. So, $\tan\left(\frac{\pi}{4} + h\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan h}{1 - \tan\left(\frac{\pi}{4}\right) \tan h}$. Since $\tan\left(\frac{\pi}{4}\right) = 1$, we have: $\tan\left(\frac{\pi}{4} + h\right) = \frac{1 + \tan h}{1 - \tan h}$ Now, substitute this back into the numerator of $f\left( \frac{\pi}{4} + h \right)$: $1 - \tan\left(\frac{\pi}{4} + h\right) = 1 - \frac{1 + \tan h}{1 - \tan h} = \frac{(1 - \tan h) - (1 + \tan h)}{1 - \tan h}$ $= \frac{1 - \tan h - 1 - \tan h}{1 - \tan h} = \frac{-2 \tan h}{1 - \tan h}$

Step 5: Simplify the denominator. The denominator is $4\left(\frac{\pi}{4} + h\right) - \pi = \pi + 4h - \pi = 4h$.

Step 6: Combine the simplified numerator and denominator and evaluate the limit. Now, the limit becomes: $\lim_{h \to 0} f\left( \frac{\pi}{4} + h \right) = \lim_{h \to 0} \frac{\frac{-2 \tan h}{1 - \tan h}}{4h}$ $= \lim_{h \to 0} \frac{-2 \tan h}{(1 - \tan h)(4h)}$ We can rearrange this expression to use the standard limit $\lim_{h \to 0} \frac{\tan h}{h} = 1$: $= \lim_{h \to 0} \left( \frac{-2}{4} \cdot \frac{\tan h}{h} \cdot \frac{1}{1 - \tan h} \right)$ $= \left( \frac{-2}{4} \right) \cdot \left( \lim_{h \to 0} \frac{\tan h}{h} \right) \cdot \left( \lim_{h \to 0} \frac{1}{1 - \tan h} \right)$ Evaluate each part:

• $\frac{-2}{4} = -\frac{1}{2}$
• $\lim_{h \to 0} \frac{\tan h}{h} = 1$
• $\lim_{h \to 0} \frac{1}{1 - \tan h} = \frac{1}{1 - \tan 0} = \frac{1}{1 - 0} = 1$

Therefore, the limit is: $= \left(-\frac{1}{2}\right) \cdot (1) \cdot (1) = -\frac{1}{2}$

Step 7: Conclude the value of $f\left( \frac{\pi}{4} \right)$. Since the function is continuous at $x = \frac{\pi}{4}$, we have $f\left( \frac{\pi}{4} \right) = \lim_{x \to \frac{\pi}{4}} f(x)$. From Step 6, we found that $\lim_{x \to \frac{\pi}{4}} f(x) = -\frac{1}{2}$. Thus, $f\left( \frac{\pi}{4} \right) = -\frac{1}{2}$.

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: While L'Hôpital's Rule can be used here, it's important to correctly differentiate the numerator and denominator. The derivative of $1 - \tan x$ is $-\sec^2 x$, and the derivative of $4x - \pi$ is $4$. Applying L'Hôpital's Rule: $\lim_{x \to \frac{\pi}{4}} \frac{-\sec^2 x}{4} = \frac{-\sec^2(\frac{\pi}{4})}{4} = \frac{-(\sqrt{2})^2}{4} = \frac{-2}{4} = -\frac{1}{2}$. This confirms our result.
• Algebraic errors with tangent addition formula: Carefully expand and simplify the expression for $\tan(\frac{\pi}{4} + h)$ and the subsequent numerator.
• Forgetting the standard limit $\lim_{h \to 0} \frac{\tan h}{h} = 1$: This is a fundamental limit for evaluating trigonometric limits and is crucial for simplifying the expression.

Summary

The problem requires us to find the value of $f(\frac{\pi}{4})$ given that $f(x)$ is continuous at $x = \frac{\pi}{4}$. By the definition of continuity, $f(\frac{\pi}{4})$ must be equal to the limit of $f(x)$ as $x$ approaches $\frac{\pi}{4}$. We evaluated this limit by substituting $x = \frac{\pi}{4} + h$, using the tangent addition formula, simplifying the expression, and employing the standard limit $\lim_{h \to 0} \frac{\tan h}{h} = 1$. This process led to the value of $-\frac{1}{2}$.

The final answer is $\boxed{-1/2}$.