<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] $$\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}}{x+1}-x$$
$$=\lim_{x\to\infty} \frac{\sqrt{x^4+x^{3}\tanh(x)+x^2}-x(x+1)}{x+1}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x+1)^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^2(x^2+2x+1)}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^4+x^{3}\tanh(x)+x^2-x^4-2x^3-x^2}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{(x+1)(\sqrt{x^4+x^{3}\tanh(x)+x^2}+x(x+1))}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{(x+1)\left(x^2\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+x^2+x\right)}$$
$$=\lim_{x\to\infty} \frac{x^{3}\tanh(x)-2x^3}{\left(x^3\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+x^3+x^2\right)+\left(x^2\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+x^2+x\right)}$$
$$=\lim_{x\to\infty} \frac{\tanh(x)-2}{\left(\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+1+\frac 1 x\right)+\left(\frac 1 x\sqrt{1+\frac  {\tanh(x)}x+\frac 1 {x^2}}+\frac 1 x+ \frac 1 {x^2}\right)} = -\frac 1 2$$
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