$$ lim _ { x \rightarrow - infty } sqrt { x ^ { 2 } + 2 x - 3 } $$
$$ lim _ { x \rightarrow - infty } sqrt { x ^ { 2 } + 2 x - 3 } $$
Answer
Explanation:
Step1: Factor out (x^{2}) from the square - root
When (x\to-\infty), (\sqrt{x^{2}}=-x). We have (\sqrt{x^{2}+2x - 3}=\sqrt{x^{2}(1+\frac{2}{x}-\frac{3}{x^{2}})}). Since (x\to-\infty), (\sqrt{x^{2}(1+\frac{2}{x}-\frac{3}{x^{2}})}=\vert x\vert\sqrt{1 + \frac{2}{x}-\frac{3}{x^{2}}}), and (\vert x\vert=-x) for (x\to-\infty). So (\sqrt{x^{2}+2x - 3}=-x\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}).
Step2: Find the limit
We want to find (\lim_{x\to-\infty}\sqrt{x^{2}+2x - 3}=\lim_{x\to-\infty}-x\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}). As (x\to-\infty), (\frac{2}{x}\to0) and (\frac{3}{x^{2}}\to0). Then (\lim_{x\to-\infty}-x\sqrt{1+\frac{2}{x}-\frac{3}{x^{2}}}=\lim_{x\to-\infty}-x\cdot1). Since (x\to-\infty), (-x\to+\infty).
Answer:
(+\infty)