For a function to be continuous at a point x=a, the limit of the function as x approaches a must equal the value of the function at that point:
\lim_{x \to \frac{\pi}{2}} f(x) = f\left(\frac{\pi}{2}\right)$ From the problem statement, we know that $f\left(\frac{\pi}{2}\right) = 3$. Therefore, we need to find $k$ such that:\lim_{x \to \frac{\pi}{2}} \frac{k\cos x}{\pi - 2x} = 3$
Evaluating the Limit using L'Hôpital's Rule:
Since substituting x=2π into the expression results in an indeterminate form of 00, we can apply L'Hôpital's Rule (differentiating the numerator and denominator with respect to x):
- The derivative of the numerator kcosx is −ksinx.
- The derivative of the denominator π−2x is −2.
Applying the rule:
x→2πlim−2−ksinx=3
\frac{-k\sin(\frac{\pi}{2})}{-2} = 3$ Since $\sin(\frac{\pi}{2}) = 1$:\frac{-k(1)}{-2} = 3\frac{k}{2} = 3k = 6$$