# Understanding what is wrong in a limit development

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?

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

You made a mistake from $$(2)$$ to $$(3)$$.
$$\sqrt{4x^2-1}=x^2\sqrt{\dfrac{4x^2-1}{x^4}}$$
so $$x \sqrt{4x^2-1}=x^2(x)\sqrt{\dfrac{4x^2-1}{x^4}}$$, not $$x^2\left(\frac1x\right)\sqrt{\dfrac{4x^2-1}{x^4}}$$
• 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. 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.