Why is $\frac{\sqrt{x^{2}-3}-2}{\sqrt{x^{2}+5}+2x}$ different from $\frac{\sqrt{1-\frac{3}{x^{2}}}-\frac{2}{x}}{\sqrt{1+\frac{5}{x^{2}}}+2}$?

Question

Based on my understanding, $\frac{\sqrt{x^{2}-3}-2}{\sqrt{x^{2}+5}+2x}$ is $\frac{\sqrt{1-\frac{3}{x^{2}}}-\frac{2}{x}}{\sqrt{1+\frac{5}{x^{2}}}+2}$ multiply by $\frac{x}{x}$, which is 1. However, when I tried to graph them, they are different when x is approaching negative infinity.

I have failed to understand the reason behind it.

Answer 1

You are assuming that $$\frac{\sqrt{x^2+5}}x=\sqrt{1+\frac5{x^2}}.$$ But that's not true when $x<0$. In that case, $$\frac{\sqrt{x^2+5}}x=-\sqrt{1+\frac5{x^2}}.$$

Answer 2

It is different for $x<0$ and moreover the second one is not defined for $x=0$.

Answer 3

Remember that $$\sqrt{x^2}=|x|, ~~~\sqrt{x^4}=x^2$$ So $$\frac{\sqrt{x^2+5}}{x}=\frac{|x|}{x} \sqrt{1+\frac{5}{x^2}}$$ So for $$x<0$$ $$\frac{\sqrt{x^2+5}}{x}=- \sqrt{1+\frac{5}{x^2}}$$ and for $$x>0$$ $$\frac{\sqrt{x^2+5}}{x}= + \sqrt{1+\frac{5}{x^2}}$$ and when $$x=0$$ you cannot divide by zero.