Question

Ma ajuta cinvea cu ex 9? Multumesc!

Answer 1

[tex]\lim_{x\rightarrow infinit} = \frac{(x+1)^2-(x-5)^2}{x^2} \frac{(x+1)^2-(x-5)^2}{x^2} = \frac{x^2+2x+1-\left(x^2-10x+25\right)}{x^2} \\=\ \textgreater \ \frac{x^2+2x+1-x^2+10x-25}{x^2} =\ \textgreater \ \frac{12x-24}{x^2} = \frac{\frac{12x}{x^2}-\frac{24}{x^2}}{\frac{x^2}{x^2}} = =\frac{12}{x}-\frac{24}{x^2}=\ \textgreater \ \\lim _{x\to \infty \:}\left(\frac{12}{x}-\frac{24}{x^2}\right) =\lim _{x\to \infty \:}\left(\frac{12}{x}\right)-\lim _{x\to \infty \:}\left(\frac{24}{x^2}\right) = 0-0 = \boxed0[/tex]