Question

Calculati urmatoarea limita
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Answer 1

Răspuns:

Explicație pas cu pas:

Este vorba despre cazul 1^inf.

$\displaystyle\lim_{x \to \infty} \left(\dfrac{x^2+3}{x^2-2x}\right)^{x+2} = \lim_{x\to\infty}\left(1 + \dfrac{2x + 3}{x^2 - 2x}\right)^{x+2} = \\ = \lim_{x\to\infty} \left[\left(1 + \dfrac{2x+3}{x^2-2x}\right)^{\frac{x^2-2x}{2x+3}}\right]^{(x+2)\cdot \frac{2x + 3}{x^2-2x}} = \lim_{x\to\infty} e^{\frac{(x+2)(2x+3)}{x^2-2x}} = \\ = \lim_{x\to\infty} e^{\frac{2x^2+4x+3x+6}{x^2-2x}} = \boxed{e^2}$