Finding the limit of ${x+\sqrt{x^2-4x+1}}$ at negative infinity

Question

I'm having trouble finding the following limit: \begin{equation} \lim_{x \to -\infty} {x+\sqrt{x^2-4x+1}} \end{equation} I tried to simplify it in many ways but couldn't get it to a form where I could evaluate the limit. How should I go about modifying this limit in order to evaluate it?

EDIT: Forgot the minus sign in front of the infinite, sorry.

Answer 1

Hint

Rewrite $$\sqrt{x^2-4x+1} = \sqrt{x^2(1-\frac 4x +\frac 1 {x^2})} = |x| \sqrt{1-\frac 4x +\frac 1 {x^2}}$$ Now, what happens inside the radical where $x$ increases more and more ?

I am sure that you can take from here.

Answer 2

$\displaystyle \lim_{x\to-\infty}x+\sqrt{x^2-4x+1}=\lim_{x\to-\infty}\frac{(x+\sqrt{x^2-4x+1})(x-\sqrt{x^2-4x+1})}{x-\sqrt{x^2-4x+1}}= \\ =\displaystyle \lim_{x\to-\infty}\frac{x^2-x^2+4x-1}{x-\sqrt{x^2-4x+1}}=\lim_{x\to-\infty}\frac{4-\frac{1}{x}}{1-\frac{|x|}{x}\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}}=\frac{4}{2}=2$

Answer 3

Set $\dfrac1x=-h$

$x^2-4x+1=\dfrac{1+4h+h^2}{h^2}\implies\sqrt{x^2-4x+1}=\dfrac{\sqrt{1+4h+h^2}}{|h|}$

$$\lim_{x \to -∞} {x+\sqrt{x^2-4x+1}}=\lim_{h\to0^+}\frac{\sqrt{1+4h+h^2}-1}h$$

$$=\lim_{h\to0^+}\frac{(1+4h+h^2)-1}{h(\sqrt{1+4h+h^2}+1)}$$

$$=\lim_{h\to0^+}\frac{4+h}{\sqrt{1+4h+h^2}+1}$$ cancelling as $h\ne0$ as $\lim_{h\to0^+}$

Now set $h$ to $0$