Question

Ajutor!!
$lim \: x -> 2 \: \: \: \: \frac{2x {}^{2} }{3x + {x}^{2} + 1}$
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Answer 1

Răspuns:

Explicație pas cu pas:

$\displaystyle\limit\lim_{x\to\infty}\dfrac{2x^2}{3x+x^2+1}=\lim_{x\to\infty}\dfrac{2x^2}{x^2\left(\dfrac{3}{x}+1+\dfrac{1}{x^2}\right)}=\boxed{2}\\\texttt{De remarcat faptul ca }\lim_{x\to\infty}\dfrac{3}{x}=\lim_{x\to\infty}\dfrac{1}{x^2}=0$