Question

Ajuda com limites no infinito.

Answer 1

$\displaystyle a)~\lim_{x\to+\infty}\left(5x^2-4x+3\right)=\infty-\infty+3=\infty\\\\b)~~\lim_{x\to\infty}10=10\\c)~~\lim_{x\to-\infty}\left(3x^4-7x^3+2x^2-5x-4\right)=\infty+\infty+\infty+\infty-4=\infty\\d)~~\lim_{x\to-\infty}\left(4-5x\right)=4+\infty=\infty\\e)~~\lim_{x\to+\infty}\sqrt{x^2-2x+2}=\sqrt{\infty}=\infty$
$\displaystyle f)~~\lim_{x\to+\infty}\frac{2}{x^2}=\frac{2}{\infty}=0\\\\g)~~\lim_{x\to-\infty} \frac{1}{x^3}=\frac{1}{-\infty}=0\\\\h)~~\lim_{x\to-\infty}\frac{x^2+4}{8x^3-1}=\lim_{x\to-\infty}\frac{\frac{1}{x}+\frac{4}{x^3}}{8-\frac{1}{x^3}}=\frac{-0-0}{8+0}=\frac{0}{8}=0$
$\displaystyle i)~~\lim_{x\to+\infty}\frac{3-2x}{5x+1}=\lim_{x\to+\infty}\frac{\frac{3}{x}-2}{5+\frac{1}{x}}=\frac{0-2}{5+0}=-\frac{2}{5}\\\\j)~~\lim_{x\to\frac{\pi}{2}}\sin x=1\\\\k)~~\lim_{x\to\frac{\pi}{2}}\cos x =0\\\\l)~~\lim_{x\to\frac{\pi}{2}}\tan x=\lim_{x\to\frac{\pi}{2}}\frac{\sin x}{\cos x}=\frac{1}{0}=\infty\\\\m)~~\lim_{t\to 0}\frac{\sin 3t}{t}=\lim_{t\to0}\frac{(\sin 3t)'}{(t)'}=\lim_{t\to0}3\cos 3t=3~~~\text{(L'H\^{o}pital)}$

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