Square root and absolute value doubt for $\lim_{x \to\ 0} \frac{{x^2+x}}{4\sqrt{x^3+x^2}}$

Question

On a math assignment question, I am trying to understand my mistake.

$$\lim_{x \to\ 0} \frac{\sqrt{x^2+x}}{4\sqrt{x^3+x^2}}$$

#EDIT 2#

I am so sorry! I wrote this out wrong when posting. Should read:

$$\lim_{x \to\ 0} \frac{{x^2+x}}{4\sqrt{x^3+x^2}}$$

#EDIT 2#

#EDIT - added my steps#

$\begin{aligned}&\frac14\lim_{x \to\ 0} \frac{{x^2+x}}{\sqrt{x^3+x^2}}\cdot\frac{\sqrt{x^3+x^2}}{\sqrt{x^3+x^2}} \\=&\frac14\lim_{x \to\ 0} \frac{x(x+1)\sqrt{x^2(x+1)}}{\sqrt{x^3+x^2}}\end{aligned}$

#EDIT - added my steps#

But when reducing this stage I had made a mistake:

$$\frac14\lim_{x \to\ 0} \frac{x(x+1)\sqrt{x^2}\sqrt{x+1}}{x^2(x+1)}$$

I pulled out the $\sqrt{x^2}$ as $x$ and not $|x|$.

Does that mistake change my answer? Do I still cross out the same way?

$\begin{aligned}&\frac14\lim_{x \to\ 0} \frac{x(x+1) |x| \sqrt{x+1}}{x^2(x+1)}\\=&\frac14\lim_{x \to\ 0} \sqrt{x+1}\\=&\frac14\sqrt{1}\\=&\frac14\end{aligned}$

So, $\lim_{x\to\ 0} f(x)$ exists

Answer 1

We have that

$$\frac{x^2+x}{4\sqrt{x^3+x^2}}=\frac{x(x+1)}{4\sqrt{x^2}\sqrt{x+1}}=\frac{x\sqrt{x+1}}{4|x|}=\frac{\sqrt{x+1}}{4}\operatorname{sign}(x)$$

and therefore

$$\lim_{x\to 0^+}\frac{\sqrt{x^2+x}}{4\sqrt{x^3+x^2}}=\lim_{x\to 0^+}\frac{\sqrt{x+1}}{4}\operatorname{sign}(x)=\frac14$$

$$\lim_{x\to 0^-}\frac{\sqrt{x^2+x}}{4\sqrt{x^3+x^2}}=\lim_{x\to 0^-}\frac{\sqrt{x+1}}{4}\operatorname{sign}(x)=-\frac14$$

Answer 2

$$\lim_{x\rightarrow0+}\frac{\sqrt{x^2+x}}{4\sqrt{x^3+x^2}}=\lim_{x\rightarrow0^+}\frac{1}{4\sqrt{x}}=+\infty.$$

For $x\rightarrow0^-$ we have $x^2+x=x(x+1)<0,$ which says that the function does not exist there.

If we say about the new limit, so for $x\rightarrow0^+$ and for $x\rightarrow0^-$ we'll get different results, which says that this limit does not exist.

Answer 3

Trading $x$ for $|x|$ may have an impact or not depending on circumstances. If used as the argument of an even function, this is harmless. If used in a minor term that does not influence the limit, this is harmless. Otherwise it is a potential mistake.