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Sa se calculeze: lim x tinde la infinit din (2x+1)(3x+1)(5x+1) supra (x+2)⁴-(x+1)⁴

Răspuns :

[tex] \lim_{x \to \infty} \frac{(2x+1)(3x+1)(5x+1)}{(x+2)^4-(x+1)^4}= \lim_{x \to \infty} \frac{x^3(2+ \frac{1}{x})(3+ \frac{1}{x})(5+ \frac{1}{x}) }{[(x+2)^2+(x+1)^2](x+2+x+1)(x+2-x-1)} [/tex]=[tex] \lim_{x \to \infty} \frac{x^3(2+ \frac{1}{x})(3+ \frac{1}{x})(5+ \frac{1}{x})}{x^3[(1+ \frac{2}{x})^2+(1+ \frac{1}{x})^2]*(2+ \frac{3}{x})*1}= \frac{2*3*5}{(1+1)*2*1}= \frac{15}{2} [/tex]