There has been 2 similar problems where I got the sign wrong so I wanted to ask you since I couldn't figure out what I was doing wrong.
The problem is: $$\lim_{x\to-\infty}\frac{\sqrt{x^2+1}}{x+1}$$
What I did was to divide both the denominator and numerator by $x$ getting: $$\lim_{x\to-\infty}\frac{\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}}{\frac{x}{x}+\frac{1}{x}}$$ Then I thought, $x^2$s would cancel out and dividing a number by infinity would give me $0$, hence: $$\lim_{x\to-\infty}\frac{\sqrt{1+0}}{1+0}=1$$
But apparently I've made a mistake and I couldn't figure out where.
The answer is $-1$.
You must write
$$\frac{|x|\sqrt{1+\frac{1}{x^2}}}{x(1+\frac{1}{x})}$$ and
$$\frac{|x|}{x}=-1$$