evaluate the following limit. use lhopitals rule when it is convenient and applicable. lim(u→π/4) (tan u…

Explanation:

Step1: Check for indeterminate form

First, substitute (u = \frac{\pi}{4}) into (\frac{\tan u-\cot u}{4u - \pi}). We have (\tan\frac{\pi}{4}=1), (\cot\frac{\pi}{4}=1), so (\lim_{u\rightarrow\frac{\pi}{4}}\frac{\tan u-\cot u}{4u - \pi}=\frac{1 - 1}{4\times\frac{\pi}{4}-\pi}=\frac{0}{0}), which is an indeterminate - form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and the denominator. The derivative of (y_1=\tan u-\cot u) is (y_1^\prime=\sec^{2}u+\csc^{2}u) (since ((\tan u)^\prime=\sec^{2}u) and ((\cot u)^\prime=-\csc^{2}u)), and the derivative of (y_2 = 4u-\pi) is (y_2^\prime = 4). So, by L'Hopital's Rule, (\lim_{u\rightarrow\frac{\pi}{4}}\frac{\tan u-\cot u}{4u - \pi}=\lim_{u\rightarrow\frac{\pi}{4}}\frac{\sec^{2}u+\csc^{2}u}{4}).

Step3: Evaluate the new limit

Substitute (u=\frac{\pi}{4}) into (\frac{\sec^{2}u+\csc^{2}u}{4}). We know that (\sec\frac{\pi}{4}=\sqrt{2}) and (\csc\frac{\pi}{4}=\sqrt{2}). Then (\sec^{2}\frac{\pi}{4}=2), (\csc^{2}\frac{\pi}{4}=2). So (\lim_{u\rightarrow\frac{\pi}{4}}\frac{\sec^{2}u+\csc^{2}u}{4}=\frac{2 + 2}{4}).

Answer:

1