Paul's Online Notes
Home / Calculus I / Applications of Derivatives / L'Hospital's Rule and Indeterminate Forms
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 4.10 : L'Hospital's Rule and Indeterminate Forms

9. Use L’Hospital’s Rule to evaluate $$\mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {x - 1} \right)\tan \left( {\displaystyle \frac{\pi }{2}x} \right)} \right]$$.

Show All Steps Hide All Steps

Start Solution

The first thing to notice here is that is not in a form that allows L’Hospital’s Rule. L’Hospital’s Rule only works on a certain class of rational functions and this is clearly not a rational function.

Note however that it is in the following indeterminate form,

${\mbox{as}}\,\,\,x \to {1^ + }\hspace{0.5in}\left( {x - 1} \right)\tan \left( {\frac{\pi }{2}x} \right) \to \left( 0 \right)\left( { - \infty } \right)$

and as we discussed in the notes for this section we can always turn this kind of indeterminate form into a rational expression that will allow L’Hospital’s Rule to be applied.

Show Step 2

The real question is do we move the first term or the second term to the denominator. At first glance it might appear that neither term will be particularly useful in the denominator. In particular, if we move the tangent to the denominator we would end up needing to differentiate a term in the form $${}^{1}/{}_{{\tan }}$$ . That doesn’t look to be all that fun to differentiate and we’re liable to end up with a mess when we are done.

However, that is exactly the term we are going to move to the denominator for reasons that will quickly become apparent.

$\mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {x - 1} \right)\tan \left( {\frac{\pi }{2}x} \right)} \right] = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1}}{{{}^{1}/{}_{{\tan \left( {\frac{\pi }{2}x} \right)}}}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1}}{{\cot \left( {\frac{\pi }{2}x} \right)}}$ Show Step 3

With a little simplification after moving the tangent to the denominator we ended up with something that doesn’t look all that bad. We’ll also see that the remainder of this problem is going to be quite simple.

Before we proceed however we should notice as well that,

${\mbox{as}}\,\,\,x \to {1^ + }\hspace{0.5in}\frac{{x - 1}}{{\cot \left( {\frac{\pi }{2}x} \right)}} \to \frac{0}{0}$

and we can use L’Hospital’s Rule on this.

Show Step 4

Applying L’Hospital’s Rule gives,

$\mathop {\lim }\limits_{x \to {1^ + }} \left[ {\left( {x - 1} \right)\tan \left( {\frac{\pi }{2}x} \right)} \right] = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x - 1}}{{\cot \left( {\frac{\pi }{2}x} \right)}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{ - \frac{\pi }{2}{{\csc }^2}\left( {\frac{\pi }{2}x} \right)}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{2}{\pi }}}$