<s>[INST] Limit of a hyperbolic trig function inside a square root I am asked to find this limit here:
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
I combined the terms to get
$$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
But if I try and factor out terms, I get
$$\lim_{x\to\infty} \frac{x^2\sqrt{1+\frac{\tanh(x)}{x}+\frac{1}{x^{2}}}-x^2(1+\frac{1}{x})}{x(1+\frac{1}{x})}$$
and that won't cancel with the x which I have on the bottom so the limit just blows up. Did I make a mistake somewhere?
The limit is apparently $\frac{-1}{2}$ but i'm not sure how that's the case. Thanks.
 [/INST] I do not knwo how much this is valid or not.
When $x\to\infty$, $\tanh(x)\to 1$. So, it seems to me that $$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x=\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}+x^2}}{x+1}-x$$ Now, $$\sqrt{x^4+x^{3}+x^2}=x^2\sqrt{1+\frac 1x+\frac 1 {x^2}}$$ Now, using $$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ and replacing $\epsilon$ by $\left(\frac 1x+\frac 1 {x^2}\right)$, $$\sqrt{1+\frac 1x+\frac 1 {x^2}}=1+\frac{1}{2 x}+\frac{3}{8 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\sqrt{x^4+x^{3}+x^2}=x^2\sqrt{1+\frac 1x+\frac 1 {x^2}}=x^2+\frac{x}{2}+\frac{3}{8}+O\left(\frac{1}{x}\right)$$ Now, using long division to get $$\frac{\sqrt{x^4+x^{3}+x^2}}{x+1}=x-\frac{1}{2}+\frac{7}{8 x}+O\left(\frac{1}{x}\right)$$ $$\frac{\sqrt{x^4+x^{3}+x^2}}{x+1}-x=-\frac{1}{2}+\frac{7}{8 x}+O\left(\frac{1}{x}\right)$$ Using $x=10$, the "exact" value is $\approx -0.422133$ while the above asymptotics gives $-\frac{33}{80}=-0.4125$
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