Evaluate the given limit \lim\limits_{x \to{\pi}{2}}{tan2x}{x-\dfrac{\pi}{2}}

\( \lim\limits_{x \to \dfrac{\pi}{2}}\dfrac{tan2x}{x-\dfrac{\pi}{2}}\)

At \( x=\dfrac{\pi}{2}\) the value of the given function takes the form \( \dfrac{0}{0}\)

Now, put So that \( x-\dfrac{\pi}{2}=y\) so that \( x\rightarrow \dfrac{\pi}{2},y\rightarrow 0\)

\( \lim\limits_{y \to 0}\dfrac{(y+\dfrac{\pi}{2})}{y}\)

\( \lim\limits_{y \to 0}\dfrac{tan2y}{y}\)

\( \lim\limits_{y \to 0}\dfrac{sin2y}{cos2y}\)

\( ( \lim\limits_{y \to 0}\dfrac{sin2y}{2y})\times ( \lim\limits_{y \to 0}\times \dfrac{2}{cos2y})\)

\( 1\times \dfrac{2}{cos0}\)

\( 1\times \dfrac{2}{1}\)

= 2