91Ó°ÊÓ

As x approaches 1 from the left, (1-x) approaches 0. So the limit of (1-x) as x approaches 1 from the left is: $$\lim_{x \rightarrow 1^{-}}(1-x) = 0$$

As x approaches 1 from the left, the argument of the tangent function approaches π/2, which is the point of discontinuity for the tangent function. Therefore, we should analyze the behavior of the tangent function as x approaches 1 from the left. This can be written as: $$\lim_{x \rightarrow 1^{-}}\tan \left(\frac{\pi x}{2}\right)$$ As x approaches 1 from the left, the angle (πx/2) approaches π/2 from the left. Since tangent is positive in the first quadrant, and its value goes to infinity as the angle approaches π/2: $$\lim_{x \rightarrow 1^{-}}\tan \left(\frac{\pi x}{2}\right)=+\infty$$

Now, we will apply the limit to the product of (1-x) and tan(πx/2). $$\lim _{x \rightarrow 1^{-}}(1-x) \tan \left(\frac{\pi x}{2}\right)$$ As x approaches 1 from the left, we've found that (1-x) approaches 0 and tan(πx/2) approaches infinity. We now have an indeterminate form: $$0 \times (+\infty)$$ To solve this, we can use L'Hôpital's Rule: $$\lim _{x \rightarrow 1^{-}}\frac{1-x}{\cot \left(\frac{\pi x}{2}\right)}$$ Differentiate (1-x) with respect to x, we get -1: $$\lim _{x \rightarrow 1^{-}}\frac{-1}{\cot \left(\frac{\pi x}{2}\right)}$$ Differentiate cot(πx/2) with respect to x, we get (-π/2)csc^2(πx/2): $$\lim _{x \rightarrow 1^{-}}\frac{-1}{-\frac{\pi}{2} \csc^2 \left(\frac{\pi x}{2}\right)}$$ Now, let's evaluate the limit as x approaches 1: $$\lim _{x \rightarrow 1^{-}}\frac{2}{\pi \csc^2 \left(\frac{\pi x}{2}\right)}$$ As x approaches 1, the cosecant function (csc) approaches infinity: $$\csc \left(\frac{\pi x}{2}\right) \rightarrow \infty$$ So, the limit of the function is: $$\lim _{x \rightarrow 1^{-}}\frac{2}{\pi \csc^2 \left(\frac{\pi x}{2}\right)}=0$$ The limit of the given function is 0.