Question

Care sunt asimptotele functiei f(x)=$\sqrt{x^{2}-4}$, cu tot ce-i sub radical in modul? 

Answer 1

[tex]\\ \sqrt{\left|x^2-4\right|} \\ \\ \left|x^2-4\right|=0 =\ \textgreater \ x^2-4=0 \\ \boxed{x=-2\quad \mathrm{sau}\quad \:x=2} \\ x\ \textless \ -2,\:-2\le \:x\ \textless \ 2,\:x\ge \:2 \\ \sqrt{\left|x^2-4\right|}\mathrm{se \:\ comporta \:\ ca}\:\ \sqrt{x^2-4}\mathrm{\:catre\:}-\infty \mathrm{\:si\:ca\:}\sqrt{x^2-4}\mathrm{\:catre\:}+\infty \\ =\ \textgreater \ \boxed{Asimptota \:\ Orizontala: y=mx+b } \\ \:\lim _{x\to -\infty \:}\frac{f\left(x\right)}{x}\:\mathrm{pentru\:\ m} \\ = \lim _{x\to \:-\infty \:}\left(\frac{\sqrt{x^2-4}}{x}\right) [/tex]

[tex]\lim _{x\to \:-\infty \:}\left(-\sqrt{1-\frac{4}{x^2}}\right) = -\sqrt{\lim _{x\to \:-\infty \:}\left(1\right)-\lim _{x\to \:-\infty \:}\left(\frac{4}{x^2}\right)} =\ \textgreater \ \\ -\sqrt{1-0} = -1 \\ \\ \lim _{x\to \:-\infty \:}\left(\sqrt{x^2-4}+1\cdot \:x\right) \\ =\lim _{x\to \:-\infty \:}\left(\frac{-4}{\sqrt{x^2-4}-1\cdot \:x}\right) \\ -4\cdot \lim \:_{x\to \:-\infty \:}\left(\frac{1}{\sqrt{x^2-4}-1\cdot \:x}\right)[/tex]

[tex]=-4\cdot \frac{\lim _{x\to \:-\infty \:}\left(1\right)}{\lim _{x\to \:-\infty \:}\left(\sqrt{x^2-4}-1\cdot \:x\right)} = -4\cdot \frac{1}{\infty \:} =0 \\ Deci \boxed{y=x,\:y=-x}[/tex]