<s>[INST] Calculating a limit or proving it does not exist I have the following function:
$$\frac{\sqrt[3]{x+3}(\sqrt{x^2+1}-\sqrt{x^2+3x+4})}{\sqrt[6]{x^2+5}}$$
The limit definition we are using is
to $-\infty$ such that $L = \lim_{x\to -\infty}f(x)$ iff 
for every $\epsilon > 0$ there exists a $M<0$ such that for every $x<M$ we have that $\lvert f(x)-L\rvert < \epsilon$.
I can see that the function is actually
$$\frac{\sqrt[3]{x+3}}{\sqrt[6]{x^2+5}}(\sqrt{x^2+1}-\sqrt{x^2+3x+4})$$
I found that
$$\lim_{x\to -\infty}\frac{\sqrt[3]{x+3}}{\sqrt[6]{x^2+5}}=1.$$
I can see that
$$(\sqrt{x^2+1}-\sqrt{x^2+3x+4})$$
has no limit but I am unable to disprove it by the definition.
After I will be able to disprove it I guess by arithmetics of limit I will be able to prove that the entire function has no limit.
Any assistance will be welcomed.
 [/INST] You have\begin{multline}\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}\left(\sqrt{x^2+1}-\sqrt{x^2+3x+4}\right)}{\sqrt[6]{x^2+5}}=\\=\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}\left(-3x-3\right)}{\sqrt[6]{x^2+5}\left(\sqrt{x^2+1}+\sqrt{x^2+3x+4}\right)}=\\=\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}}{\sqrt[6]{x^2+5}}\lim_{x\to-\infty}\frac{-3x-3}{\sqrt{x^2+1}+\sqrt{x^2+3x+4}}=\\=(-1)\times\frac32=-\frac32.\end{multline}We have$$\lim_{x\to-\infty}\frac{x+1}{\sqrt{x^2+1}+\sqrt{x^2+3x+4}}=-\frac12$$because\begin{align}\lim_{x\to-\infty}\frac{x+1}{\sqrt{x^2+1}+\sqrt{x^2+3x+4}}&=\lim_{x\to-\infty}\frac{1+\frac1x}{-\sqrt{1+\frac1{x^2}}-\sqrt{1-\frac3x-\frac4{x^2}}}\\&=\frac1{-1-1}\\&=-\frac12.\end{align}
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