0
$\begingroup$

So, I am having trouble trying to calculate

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

Can you guys give me a hint? Thank you.

  • 0
    $$(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)=?$$ OR set $x=\cot2y$ – 2017-02-10
  • 0
    = (√(x^2+1))^2-x^2= x^2+1-x^2=1 that will never give me + infinite – 2017-02-10
  • 0
    wait I saw my mistake -(-infinite)= + infinite. Sorry, that was so simple... – 2017-02-10
  • 0
    That's it! Thank you. – 2017-02-10

4 Answers 4

2

For $x < 0$ $$ \sqrt{x^2+1} - x = \sqrt{x^2+1} + |x| > 2 |x| $$

  • 0
    That's it! Thank you. – 2017-02-10
0

Since $x<0$ we have

$$+\infty = \lim_{x\to -\infty}|x|\le\lim_{x\to-\infty}\sqrt{x^2+1}\le\lim_{x\to-\infty}\sqrt{x^2+1}-x$$

  • 0
    That's it! Thank you. – 2017-02-10
  • 0
    @GiuR my pleasure. – 2017-02-10
0

Obviously $$ \sqrt{x^2+1}-x>-x $$ What is $\lim\limits_{x\to-\infty}(-x)$?

Alternatively, observe this is not an “indeterminate form”, because it is $\sqrt{x^2+1}+(-x)$ and both summands have limit $\infty$.

Maybe your problem stems from trying to pull $x$ outside the square root, doing $$ \sqrt{x^2+1}=x\sqrt{1+\frac{1}{x^2}} $$ but this is a big mistake: for $x<0$, the left-hand side is positive, but the right-hand side is negative, so they cannot be equal. Remember that $\sqrt{x^2}=|x|$, so, for $x<0$ (which you can assume, since you have a limit for $x\to-\infty$), $$ \sqrt{x^2+1}=|x|\sqrt{1+\frac{1}{x^2}}=-x\sqrt{1+\frac{1}{x^2}} $$ Hence you can rewrite your limit as $$ \lim_{x\to-\infty}-x\left(\sqrt{1+\frac{1}{x^2}}+1\right) $$ where the term in parentheses has limit $2$, so we have $-(-\infty)\cdot2=\infty$.

0

$$\lim_{x \to -\infty} \sqrt{x^2+1}-x = \lim_{y \to \infty} \sqrt{(-y)^2+1}-(-y)=\lim_{y \to \infty} \sqrt{y^2+1}+y=\infty+\infty=\infty$$