[tex]\lim_{x \to +\infty} [ln(e^x+2^x)-x+x-\sqrt{x^2-4x+1} ]=\\\\ \lim_{x \to +\infty} [ln(e^x+2^x)-lne^x+x-\sqrt{x^2-4x+1} ]=\\\\ \lim_{x \to +\infty} [ln\frac{e^x+2^x}{e^x} +\frac{(x-\sqrt{x^2-4x+1})( x+\sqrt{x^2-4x+1})}{x+\sqrt{x^2-4x+1}} ]=\\\\ \lim_{x \to +\infty}[ln(1+(\frac{2}{e} )^x)+\frac{x^2-x^2+4x-1}{x+\sqrt{x^2-4x+1}} ]=\\\\ \lim_{x \to +\infty}(ln1+\frac{4x-1}{x+\sqrt{x^2-4x+1}} )=\\\\ \lim_{x \to +\infty}\frac{x(4-\frac{1}{x}) }{x(1+\sqrt{1-\frac{4}{x}+\frac{1}{x^2} }) } =\frac{4}{2}=2[/tex]
Raspuns: c) 2
Un alt exercitiu cu limite gasesti aici: https://brainly.ro/tema/957231
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