How to compute $\lim _{x\to \infty }\left(2\sqrt{x^2-4}-2x+1\right)$? I have a limit that I don't know how to compute: $$ \lim _{x\to \infty }\left(2\sqrt{x^2-4}-2x+1\right) $$
What I've tried : $$ \lim _{x\to \infty }\left(2\sqrt{x^2-4}-2x+1\right) = \lim _{x\to \infty }\left(x\left(2\sqrt{1-\frac{4}{x^2}}-2+\frac{1}{x}\right)\right) = \lim _{x\to \infty }\left(\frac{\left(2\sqrt{1-\frac{4}{x^2}}-2+\frac{1}{x}\right)^{'}}{\left(\frac{1}{x}\right)^{'}}\right) = \lim _{x\to \infty }\left(\frac{\left(\frac{1}{\sqrt{1-\frac{4}{x^2}}}+-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}\right) = \lim _{x\to \infty }\left(\left(-\frac{x^2}{\sqrt{1-\frac{4}{x^2}}}+1\right)\right) = -\infty $$ But the real answer is $1$. Where am I wrong? Thanks for any response.
As $\sqrt{x^2-4}-x=\frac{x^2-4-x^2}{\sqrt{x^2-4}+x}=\frac{-4}{\sqrt{x^2-4}+x}$ tends to $0$ when x tends to infinity, we have $$\lim_{x\to\infty} \sqrt{x^2-4}-x+1=\lim_{x\to\infty} \frac{-4}{\sqrt{x^2-4}+x}+1=0+1=1.$$
Hint $2\sqrt{x^2 - 4} \to 2\sqrt{x^2} \to 2x$ as $x \to \infty$, simplifying your expression $\underbrace{2\sqrt{x^2 - 4}}_{\to 2x} - 2x + 1$ to...?