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.
 A: The limit cannot be found in your way.
Multiplying $$\frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
by
$$\frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1)}{\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1)}\ (=1)$$
gives
$$\frac{(x^4+x^3\tanh(x)+x^2)-(x^2+x)^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\frac{x^3(\tanh(x)-2)}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\frac{\tanh(x)-2}{(1+\frac 1x)\left(\sqrt{1+\frac{\tanh(x)}{x}+\frac{1}{x^2}}+1+\frac 1x\right)}\to -\frac 12$$
as $x\to\infty$.
A: $$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
$$=\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x+1)^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x^2+2x+1)}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^4-2x^3-x^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{(x+1)\left(x^2\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+x^2+x\right)}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{\left(x^3\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+x^3+x^2\right)+\left(x^2\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+x^2+x\right)}$$
$$=\lim_{x\to\infty} \frac{\tanh(x)-2}{\left(\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+1+\frac 1 x\right)+\left(\frac 1 x\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+\frac 1 x+ \frac 1 {x^2}\right)} = -\frac 1 2$$
A: 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 $-\frac{33}{80}=-0.4125$
