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,

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