Question

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}(x - 1)\tan\frac{\pi  x}{2}, \text{ if } & x \neq 1 \\ k , if & x = 1\end{cases}\] at x = 1at x = 1

Answer 1

Given 

\[f\left( x \right) = \binom{\left( x - 1 \right)\tan\frac{\pi x}{2}, \text{ if }  x \neq 1}{k, \text{ if }  x = 1}\] If f(x) is continuous at x = 1, then 

\[\lim_{x \to 1} f\left( x \right) = f\left( 1 \right)\]
\[ \Rightarrow \lim_{x \to 1} \left( x - 1 \right) \tan\frac{\pi x}{2} = k\]

\[\text{ Putting }  x - 1 = y, \text{ we get } \]

\[\lim_{y \to 0} y \tan\frac{\pi\left( y + 1 \right)}{2} = k\]
\[ \Rightarrow \lim_{y \to 0} y \tan\left( \frac{\ piy}{2} + \frac{\pi}{2} \right) = k\]
\[ \Rightarrow \lim_{y \to 0} y \tan\left( \frac{\pi}{2} + \frac{\ piy}{2} \right) = k\]
\[ \Rightarrow - \lim_{y \to 0} y \cot\left( \frac{\ piy}{2} \right) = k\]
\[ \Rightarrow \frac{- 2}{\pi} \lim_{y \to 0} \frac{\frac{\ piy}{2}\cos\left( \frac{\ piy}{2} \right)}{\sin\left( \frac{\ piy}{2} \right)} = k\]
\[ \Rightarrow \frac{- 2}{\pi} \frac{\lim_{y \to 0} \cos\left( \frac{\ piy}{2} \right)}{\lim_{y \to 0} \left( \frac{\sin\left( \frac{\ piy}{2} \right)}{\frac{\ piy}{2}} \right)} = k\]
\[ \Rightarrow \frac{- 2}{\pi} \times \frac{1}{1} = k\]
\[ \Rightarrow k = \frac{- 2}{\pi}\]