<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] 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 $x\to\infty$.
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