Question

Am nevoie de rezolvare completă… răspunsul este A

Answer 1

Explicație pas cu pas:

rezolvare cu Regula lui l'Hospital:

$lim_{x - > 0}( \frac{x \sin(x) }{{e}^{x} + {e}^{ - x} - 2}) \\ = lim_{x - > 0}( \frac{x \sin(x) }{ \frac{{e}^{2x} + 1 - 2{e}^{x}}{ {e}^{x} } }) = lim_{x - > 0}( \frac{ {e}^{x} x \sin(x) }{({e}^{x} - 1)^{2} }) \\ = lim_{x - > 0}( {e}^{x}) \times lim_{x - > 0}( \frac{x \sin(x) }{({e}^{x} - 1)^{2} }) \\ = 1 \times lim_{x - > 0}( \frac{x \cos(x) + \sin(x) }{2 {e}^{x} ({e}^{x} - 1) }) \\ = lim_{x - > 0}(\frac{{e}^{ - x}(x \cos(x) + \sin(x)) }{2({e}^{x} - 1) }) \\ = \frac{1}{2}lim_{x - > 0}({e}^{ - x}) \times lim_{x - > 0}(\frac{x \cos(x) + \sin(x)}{{e}^{x} - 1}) \\ = \frac{1}{2} \times lim_{x - > 0}(\frac{x \cos(x)}{{e}^{x} - 1} + \frac{\sin(x)}{{e}^{x} - 1}) \\ = \frac{lim_{x - > 0}(\frac{x \cos(x)}{{e}^{x} - 1}) + lim_{x - > 0}(\frac{\sin(x)}{{e}^{x} - 1})}{2} \\ = \frac{lim_{x - > 0}(\cos(x)) \times lim_{x - > 0}(\frac{x}{{e}^{x} - 1}) + lim_{x - > 0}(\frac{\sin(x)}{{e}^{x} - 1})}{2} \\ = \frac{1 \times lim_{x - > 0}(\frac{1}{{e}^{x}}) + lim_{x - > 0}(\frac{\cos(x)}{{e}^{x}})}{2} \\ = \frac{1 + 1}{2} = 1$