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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?

Thanks in advance.

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    $\begingroup$ Limit is $-1$ not $1$ so you'd get $-\infty\cdot 0$ $\endgroup$
    – David P
    Commented Oct 13, 2020 at 11:16

1 Answer 1

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Note that by $x=-y$

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

which leads to an indeterminate form.

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