Odpowiedź:
[tex]\Large\boxed{\lim\limits_{x\to-\infty}\left(x-\sqrt{x^2+2x-3}\right)=-\infty}[/tex]
Szczegółowe wyjaśnienie:
[tex]\lim\limits_{x\to-\infty}\left(x-\sqrt{x^2+2x-3}\right)=\\[5]\lim\limits_{x\to-\infty}\left(x-|x|\cdot\sqrt{1+\dfrac{2}{x}-\dfrac{3}{x^2}}\right)=\\[5]\lim\limits_{x\to-\infty}x-\lim\limits_{x\to-\infty}\left(|x|\cdot\sqrt{1+\dfrac{2}{x}-\dfrac{3}{x^2}}\right)=\\[5]-\infty-\infty\cdot 1=-\infty[/tex]
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Odpowiedź:
[tex]\Large\boxed{\lim\limits_{x\to-\infty}\left(x-\sqrt{x^2+2x-3}\right)=-\infty}[/tex]
Szczegółowe wyjaśnienie:
[tex]\lim\limits_{x\to-\infty}\left(x-\sqrt{x^2+2x-3}\right)=\\[5]\lim\limits_{x\to-\infty}\left(x-|x|\cdot\sqrt{1+\dfrac{2}{x}-\dfrac{3}{x^2}}\right)=\\[5]\lim\limits_{x\to-\infty}x-\lim\limits_{x\to-\infty}\left(|x|\cdot\sqrt{1+\dfrac{2}{x}-\dfrac{3}{x^2}}\right)=\\[5]-\infty-\infty\cdot 1=-\infty[/tex]