Pregunta: limx→-∞5x4-3x2+1=limx→∞5x4-3x2+1=limx→-∞-2x3-x2+1= limx→∞-2x3-x2+1=limx→-∞2x+65x= limx→-∞1x2-9=limx→∞(1+x)(2+x)(1+2x)(2+3x)=limx→-∞x2+4x24x+1=

$\underset{x\to -\infty }{lim}5x^4-3x^2+1=$

$\underset{x\to \infty }{lim}5x^4-3x^2+1=$

$\underset{x\to -\infty }{lim}-2x^3-x^2+1=\underset{x\to \infty }{lim}-2x^3-x^2+1=$

$\underset{x\to -\infty }{lim}\frac{2x+6}{5x}=\underset{x\to -\infty }{lim}\frac{1}{x^2-9}=$

$\underset{x\to \infty }{lim}\frac{(1+x)(2+x)}{(1+2x)(2+3x)}=$

$\underset{x\to -\infty }{lim}\frac{\sqrt[\phantom{2}]{x^2+4x}}{4x+1}=$