Question

Lim 1 + cos TX5.x>1 tanưrex​

Answer 1

Answer: 1/2

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

Step-by-step explanation: