Chapter 9.2, Problem 13E

To determine

To Find: The value of a limit, both graphically and using L'Hôpital's Rule, after identifying the indeterminate form.

Answer to Problem 13E

The value of the limit is $-1$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to \pi }{\lim }\frac{\csc x}{1+\cot x}$

Method:

Graphically: : Plot the graph of the expression in $x$ and identify the value to which the expression approaches as $x$ approaches to $\pi$ . $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to a,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation:

(i)The graph of the expression is

  

In the above graph, the value of the expression approaches to $-1$ when $x$ approaches to $\pi$ .

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

(ii). As $a\to {\pi }^{-}$ ,

  $\csc a\to \infty$ and $1+\cot a\to -\infty$

Therefore, the limit is of the indeterminate form $\frac{\infty }{-\infty }$ .

And as $a\to {\pi }^{+}$ ,

  $\csc a\to -\infty$ and $1+\cot a\to \infty$

Therefore, the limit is of the indeterminate form $\frac{-\infty }{\infty }$ .

So, apply L'Hôpital's Rule to write

  $\begin{array}{l}\underset{x\to \pi }{\lim }\left(\frac{\csc x}{1+\cot x}\right)=\underset{x\to \pi }{\lim }\left(\frac{-\csc x\cot x}{-{\csc }^{2}x}\right)=\underset{x\to \pi }{\lim }\left(\sec x\right)=\sec \pi =-1\\ \end{array}$

The limit determined using both the methods is indeed the same.