Question

1–38 ■ Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

\(\mathop {\lim }\limits_{x \to {1^ + }} \ln x\tan (\pi x/2)\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to {1^ + }} \ln x\tan \left( {\frac{{\pi x}}{2}} \right)\) is \( - \frac{2}{\pi }\).

Step-by-step solution

Given information

The given function is \(\mathop {\lim }\limits_{x \to {1^ + }} \ln x\tan \left( {\frac{{\pi x}}{2}} \right)\).

Concept of L' Hospital's Rule

Suppose that one of the following cases.

\(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{0}{0}\quad {\rm{ OR }}\quad \mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{ \pm \infty }}{{ \pm \infty }}\)

Here\(a\)can be any real number, infinity, or negative infinity.

In these cases,\(\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{{f^\prime }(x)}}{{{g^\prime }(x)}}\).

Intermediate form of given function

Express given function as, \(\mathop {\lim }\limits_{x \to {1^ + }} \ln x\tan \left( {\frac{{\pi x}}{2}} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\ln x \cdot \sin \left( {\frac{{\pi x}}{2}} \right)}}{{\cos \left( {\frac{{\pi x}}{2}} \right)}}\).

Obtain the value of the function as \(x\) approaches \({1^ + }\).

As \(x\) approaches \(\infty \), the numerator \(\ln x \cdot \sin \left( {\frac{{\pi x}}{2}} \right) = \ln (1) \cdot \sin \left( {\frac{\pi }{2}} \right)\) approaches to \(0\) and denominator \(\cos \left( {\frac{{\pi x}}{2}} \right) = \cos \left( {\frac{\pi }{2}} \right)\) also approaches to \(0\).

Thus, \(\mathop {\lim }\limits_{x \to {1^ + }} \ln x\tan \left( {\frac{{\pi x}}{2}} \right) = \frac{0}{0}\) is in an indeterminate form.

  Step 4: Apply L’ Hospital’s Rule and simplify the terms

Therefore, apply L’ Hospital’s Rule and obtain the limit.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {1^ + }} \ln x\tan \left( {\frac{{\pi x}}{2}} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\ln x \cdot \sin \left( {\frac{{\pi x}}{2}} \right)}}{{\cos \left( {\frac{{\pi x}}{2}} \right)}}\\ = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\frac{\pi }{2} \cdot \cos \left( {\frac{{\pi x}}{2}} \right) \cdot \ln x + \sin \left( {\frac{{\pi x}}{2}} \right) \cdot \frac{1}{x}}}{{ - \frac{\pi }{2} \cdot \sin \left( {\frac{{\pi x}}{2}} \right)}}\\ = - \frac{{\frac{\pi }{2} \cdot \cos \left( {\frac{\pi }{2}} \right) + \sin \left( {\frac{\pi }{2}} \right) \cdot \frac{1}{1}}}{{\frac{\pi }{2} \cdot \sin \left( {\frac{\pi }{2}} \right)}}\end{array}\)

Simplify the terms further.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {1^ + }} \ln x\tan \left( {\frac{{\pi x}}{2}} \right) = - \frac{{\frac{\pi }{2} \cdot \cos \left( {\frac{\pi }{2}} \right) + \sin \left( {\frac{\pi }{2}} \right) \cdot \frac{1}{1}}}{{\frac{\pi }{2} \cdot \sin \left( {\frac{\pi }{2}} \right)}}\\ = - \frac{{\frac{\pi }{2} \cdot 0 + 1}}{{\frac{\pi }{2} \cdot 1}}\\ = - \frac{1}{{\frac{\pi }{2}}}\\ = - \frac{2}{\pi }\end{array}\)

Thus, the limit of the function is \( - \frac{2}{\pi }\).