Chapter 9.2, Problem 37E

To determine

To Find: The value of a limit using L'Hôpital's Rule.

Answer to Problem 37E

The value of the limit is $1$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{y\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\left(\frac{\pi }{2}-y\right)\tan y$ .

Method: $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to a,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: As $y\to {\frac{\pi }{2}}^{-}$ ,

  $\frac{\pi }{2}-y\to 0$ and $\tan y\to \infty$

Therefore, the limit is of the indeterminate form $0\cdot \infty$ . So, since $\cot \frac{\pi }{2}=0$ ,write $\tan y$ as $\frac{1}{\cot y}$ to convert the limit into $\frac{0}{0}$ form as below

  $\underset{y\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\left(\frac{\pi }{2}-y\right)\tan y=\underset{y\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\frac{\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-y\right)}{\cot y}$

Now, apply L'Hôpital's Rule to write

  $\begin{array}{l}\underset{y\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\frac{\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-y\right)}{\cot y}=\underset{y\to \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\lim }\left(\frac{0-1}{-{\csc }^{2}y}\right)\\ =\frac{-1}{-{\csc }^2\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{-1}{-1}=1\end{array}$

Therefore, the value of the limit is $1$ .