Question

$$ lim _ { x ightarrow - infty } sqrt { x ^ { 2 } + 2 x - 3 } $$

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:

\(+\infty\)