# Contradictory limit values when using two different ways.

I am trying to calculate the limit for this function and it gives 2 different answers when I used 2 different methods. I hope you will tell me why this happened.

The function: $$f(x) =$$ $$x$$ + $$\sqrt{x^2+4}$$.

Since the first substitution gives an indeterminate form $$-\infty +\infty$$, I tried first to take $$x$$ as a common factor. $$f(x) = x( 1+ \frac{\sqrt{x^2+4}}{x})$$

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

So,

$$\lim_{x \to -\infty} f(x)= ( -\infty)(2)= - \infty$$.

However, if I used rationalization:

$$f(x) = \frac{(x+ \sqrt{x^2+4}) (x- \sqrt{x^2+4})} {x- \sqrt{x^2+4}}$$.

$$\lim_{x \to -\infty} f(x)= \lim_{x \to -\infty} \frac {-4}{x-\sqrt{x^2+4}}= \frac{-4} {-\infty}=0$$.

According to my book, rationalization method is the correct one.

But I need to know what is wrong in my first method, and when must I use rationalization instead of any other method?

• Limit is $-1$ not $1$ so you'd get $-\infty\cdot 0$ Commented Oct 13, 2020 at 11:16
Note that by $$x=-y$$
$$\lim_{x \to -\infty}\frac{\sqrt{x^2}} {x} =\lim_{y \to \infty}\frac{\sqrt{y^2}} {-y} =-1$$