I have the following limit:

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

I know that the result is $-2$ and I know how to achieve it. However on the first try I made the following development and I still can't see what I am doing wrong:

$$\mathbf1)\lim_{x\to-\infty} \frac{\sqrt {4x^2-1}}{x}$$

$$\mathbf2)\lim_{x\to-\infty} \frac{4x^2-1}{x\sqrt{4x^2-1}}$$

$$\mathbf3)\lim_{x\to-\infty} \frac{x^2(4-\frac{1}{x^2})}{x^2(\frac{1}{x})\sqrt{\frac{4x^2-1}{x^4}}}$$

$$\mathbf4)\lim_{x\to-\infty} \frac{(4-\frac{1}{x^2})}{(\frac{1}{x})\sqrt{\frac{4}{x^2}-\frac{1}{x^4}}}$$

Denominator goes to zero and I remain with $\frac{4}{0}= \infty$

Where is the mistake?

  • 1
    $\begingroup$ Why did you switch $x$ to $\frac1x$? $\endgroup$ Jul 13, 2020 at 15:04
  • 1
    $\begingroup$ What is the meaning of $x̧$? You seem to be using $x$ and $x̧$ interchangeably. $\endgroup$
    – Théophile
    Jul 13, 2020 at 15:08
  • 1
    $\begingroup$ The denominator of $3)$ is not equal to the denominator of $2)$. look at the radicand of $3)$ $\endgroup$
    – Vajra
    Jul 13, 2020 at 15:09

2 Answers 2


You made a mistake from $(2)$ to $(3)$.


so $x \sqrt{4x^2-1}=x^2(x)\sqrt{\dfrac{4x^2-1}{x^4}}$, not $x^2\left(\frac1x\right)\sqrt{\dfrac{4x^2-1}{x^4}}$

  • 1
    $\begingroup$ Sure, thank you! I was simplifying in my head the $x$ as $x^2(\frac{1}{x})$ without realizing the $x^2$ was already there because of the square root simplification. $\endgroup$
    – q0mlm
    Jul 13, 2020 at 15:30

Between steps 2 and 3, you factored out $x^2$ both from $x$ and from $\sqrt{4x^2 - 1}$ in the denominator. You should have ended up with $x^2 \left(\frac{1}{x}\right)\sqrt{4x^2 - 1}$ or $x^2(x)\sqrt{\frac{4x^2 - 1}{x^4}}$, but taking the $x^2$ out of both means you should have a factor of $x^4$ instead.


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