Correct Answer - Option 4 :
\(-\frac{a}{\pi}\)
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)
II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{∞ }{∞ }\)
Then we can apply L-Hospital Rule
\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
Indeterminate form of the type infinity divided by infinity (∞/∞) or zero divided by zero (0/0)
Calculation:
Given:
\(X=\mathop {\lim }\limits_{y \to a} (\sin \frac{{y - a}}{2}\tan \frac{{\pi y}}{{2a}})\) ---(1)
Put y = a in equation 1)
X = 0 × ∞
It is not an indeterminate form so 1st convert it into indeterminate form;
\(X=\mathop {\lim }\limits_{y \to a} \frac{{\sin \frac{{y - a}}{2}}}{{\cot \frac{{\pi y}}{{2a}}}}\)
Now put y = a;
\(X=\frac{0}{0}\)
Now, we can apply the L-Hospital rule;
\(\mathop {\lim }\limits_{y \to a} \frac{{\frac{1}{2}\cos \left( {\frac{{y - a}}{2}} \right)}}{{\frac{{ - \pi }}{{2a}} \cdot cose{c^2}\frac{{ny}}{{2a}}}} = - \frac{a}{\pi }\)