1. Stating the problem: Evaluate the limit $$\lim_{x\to -1} \arctan\left(\frac{2x}{1-x^2}\right)$$. 2. Understand the expression inside the arctan function: $$\frac{2x}{1-x^2}$$. 3. Factor the denominator: $$1 - x^2 = (1-x)(1+x)$$. 4. Note that as $$x \to -1$$, $$1+x \to 0$$, so the denominator approaches zero. 5. Approach the limit from the left side $$x \to -1^-$$: For values just less than -1, $$1+x < 0$$ and $$1-x > 0$$, so denominator is negative. Numerator $$2x$$ approaches $$-2$$ (negative). 6. Therefore, the fraction $$\frac{2x}{1-x^2}$$ as $$x \to -1^-$$ tends to $$\frac{-2}{0^-} = +\infty$$. 7. Approach the limit from the right side $$x \to -1^+$$: For values just greater than -1, $$1+x > 0$$ and $$1-x > 0$$, denominator positive and numerator negative close to -2, so fraction $$\to \frac{-2}{0^+} = -\infty$$. 8. Since the limits from left and right of the inside function $$\frac{2x}{1-x^2}$$ go to $$+\infty$$ and $$-\infty$$ respectively, the inside expression is not approaching a single value. 9. Evaluate the limit of the overall function using continuity and definition of arctan: $$\lim_{x\to -1^-} \arctan\left(\frac{2x}{1-x^2}\right) = \arctan(+\infty) = \frac{\pi}{2}$$ $$\lim_{x\to -1^+} \arctan\left(\frac{2x}{1-x^2}\right) = \arctan(-\infty) = -\frac{\pi}{2}$$ 10. Since the left-hand and right-hand limits differ, the limit does not exist. Final answer: The limit $$\lim_{x\to -1} \arctan\left(\frac{2x}{1-x^2}\right)$$ does not exist because the left and right limits are different.