Question

Dúvida sobre limites
Lim senx
x->π/4


Lim. 2x^2+x^2+4/ -x^2-2x+1
x->infinito​

Answer 1

$\lim_{x\to{\frac{\pi}{4}}}(\sin(x)) \\ = \sin( \frac{\pi}{4}) \\ = \frac{ \sqrt{2} }{2}$

$\lim_{x\to\infty}\frac{2x^2 + x^2 + 4}{-x^2 - 2x + 1}$

$=\lim_{x\to\infty}\frac{3x^2 + 4}{-x^2 - 2x + 1}$

$=\lim_{x\to\infty}\frac{x^2(3 + \frac{4}{x^2})}{x^2(-1 - \frac{2}{x} + \frac{1}{x^2})}$

$</p><p>=\lim_{x\to\infty}\frac{3 + \frac{4}{x^2} }{-1 - \frac{2}{x} + \frac{1}{x^2}} \\ = \frac{3 + \frac{4}{ \infty} }{ - 1 - \frac{2}{ \infty} + \frac{1}{ \infty } } \\ = \frac{3 + 0}{ - 1 - 0 + 0} \\ = - 3$