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

Answer:

$\lim_{u\rightarrow\frac{\pi}{4}}\frac{\sec^{2}u+\csc^{2}u}{1}$

Explanation:

Step1: Check indeterminate form

When $u = \frac{\pi}{4}$, $\tan u-\cot u=\tan\frac{\pi}{4}-\cot\frac{\pi}{4}=1 - 1=0$ and $u-\frac{\pi}{4}=0$. So it's in $\frac{0}{0}$ form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $y=\tan u-\cot u$ is $y'=\sec^{2}u+\csc^{2}u$ (using $\frac{d}{du}\tan u=\sec^{2}u$ and $\frac{d}{du}\cot u=-\csc^{2}u$), and the derivative of $y = u-\frac{\pi}{4}$ is $y' = 1$. So by L'Hopital's Rule, $\lim_{u\rightarrow\frac{\pi}{4}}\frac{\tan u-\cot u}{u - \frac{\pi}{4}}=\lim_{u\rightarrow\frac{\pi}{4}}\frac{\sec^{2}u+\csc^{2}u}{1}$.