Let f : (0, 2) $\to$ R be defined as f(x) = log₂$\left( {1 + \tan \left( {{{\pi x} \over 4}} \right)} \right)$. Then, $$\mathop {\lim }\limits_{n \to \infty } {2 \over n}\left( {f\left( {{1 \over n}} \right) + f\left( {{2 \over n}} \right) + ... + f(1)} \right)$$ is equal to ___________.

$$E = 2\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} f\left( {{r \over n}} \right)$$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {{\pi x} \over 4}} \right)dx} $$ ..... (i)<br><br>replacing x $\to$ 1 $-$ x<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {\pi \over 4}(1 - x)} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan \left( {{\pi \over 4} - {\pi \over 4}x} \right)} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + {{1 - \tan {\pi \over 4}x} \over {1 + \tan {\pi \over 4}x}}} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( { {2 \over {1 + \tan {{\pi x} \over 4}}}} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\left( {\ln 2 - \ln \left( {1 + \tan {{\pi x} \over 4}} \right)} \right)dx} $$ ..... (ii)<br><br>equation (i) + (ii)<br><br>2E = 2 $\Rightarrow$ E = 1