limite

PedroCunha escreveu:Por L'Hopital:

\\ \lim_{x \to 1} \frac{1+\cos (\pi \cdot x)}{\tan^2 (\pi \cdot x)} = \lim_{x \to 1} \frac{-\pi \cdot \sin (\pi \cdot x)}{\pi \cdot 2 \tan (\pi \cdot x) \cdot  \sec^2 (\pi \cdot x)} \\\\ = \frac{1}{2} \cdot \lim_{x \to 1} \frac{-\sin (\pi \cdot x) \cdot \cos^2 (\pi \cdot x)}{\tan (\pi \cdot x)}

Aplicando L'Hopital novamente:

\frac{1}{2} \cdot \lim_{x \to 1} \frac{-\sin (\pi \cdot x) \cdot \cos^2 (\pi \cdot x)}{\tan (\pi \cdot x)}\\\\ = \frac{1}{2} \cdot \lim_{x \to 1} \cdot \frac{-(\pi \cdot \cos(\pi \cdot x) \cdot \cos^2 (\pi \cdot x) + \sin(\pi \cdot x) \cdot (2 \cos(\pi \cdot x) \cdot (-\sin (\pi \cdot x) \cdot \pi)}{\sec^2 (\pi \cdot x) \cdot \pi} \\\\ = \frac{1}{2} \cdot \lim_{x \to 1} \frac{-\pi \cdot \cos (\pi \cdot x) \cdot (\cos^2(\pi \cdot x) - 2 \cdot sin^2(\pi \cdot x))}{\sec^2(\pi \cdot x) \cdot \pi} \\\\ = \frac{1}{2} \cdot \lim_{x \to 1} \cos^3( \pi \cdot x) \cdot (-\cos^2(\pi \cdot x) + 2 \cdot \sin^2(x \cdot \pi) = \frac{1}{2} \cdot (-1 \cdot (-1 + 0)) = \frac{1}{2}

Att.,
Pedro