\( \lim _ { x \rightarrow \pi / 4 } \frac { x } { x - \pi / 4 } \int _ { \pi / 4 } ^ { x } \frac { \cos ( t ) } { t } d t \)

To evaluate the limit \[ \lim _ { x \rightarrow \pi / 4 } \frac { x } { x - \pi / 4 } \int _ { \pi / 4 } ^ { x } \frac { \cos ( t ) } { t } d t, \] we begin by recognizing that as \( x \) approaches \( \pi / 4 \), both the numerator and the denominator approach 0, which suggests we can use L'Hôpital's rule. First, define the integral: \[ f(x) = \int_{\pi/4}^{x} \frac{\cos(t)}{t} dt. \] Now, we rewrite the limit: \[ L = \lim_{x \to \pi/4} \frac{x f(x)}{x - \pi/4}. \] Since \( f(\pi/4) = 0 \), we can apply L'Hôpital's rule: \[ L = \lim_{x \to \pi/4} \frac{f(x) + x f'(x)}{1}. \] Next, we need to compute \( f'(x) \): \[ f'(x) = \frac{\cos(x)}{x}. \] Now substitute \( x = \pi/4 \): 1. Evaluate \( f(x) \) as \( x \rightarrow \pi/4 \): \[ f(\pi/4) = 0. \] 2. Now, find \( f'(\pi/4) \): \[ f'(\pi/4) = \frac{\cos(\pi/4)}{\pi/4} = \frac{\sqrt{2}/2}{\pi/4} = \frac{2\sqrt{2}}{\pi}. \] Thus, substituting back into our limit: \[ L = \lim_{x \to \pi/4} \left( f(x) + x f'(x) \right) = 0 + \frac{\pi/4 (2\sqrt{2}/\pi)}{1} = \frac{\sqrt{2}}{2}. \] Putting it all together, we find our final result: \[ \lim _ { x \rightarrow \pi / 4 } \frac { x } { x - \pi / 4 } \int _ { \pi / 4 } ^ { x } \frac { \cos ( t ) } { t } d t = \frac{\sqrt{2}}{2}. \]