<s>[INST] Is limits $\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$? $$\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$$
I used an online calculator and it said it was actually $=\infty$
Here's how I calculate it:
$$\lim_{x\to\infty}\dfrac{x^2+4-4x}{2x+1}=\lim_{x\to\infty}\dfrac{({x\over x}-{2\over x})({x\over x}-{2\over x})}{({2x\over x}+{1\over x})}=\dfrac{(1-0)(1-0)}{(2+0)}={1 \over 2}$$
 [/INST] You have accidentally divided the numerator by $x^2$ and the denominator only by $x$. For the numerator we get
$$(x-2)^2 = \left[x\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\right]\cdot \left[x\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\right] = x^2\left(1-\tfrac{2}{x}\right)^2,$$
so the fraction is
$$\frac{(x-2)^2}{2x+1} = \frac{x^2\left(1-\tfrac{2}{x}\right)^2}{x\left(2+\frac{1}{x}\right)} = \underbrace{x}_{\to\infty}\underbrace{\frac{\left(1-\tfrac{2}{x}\right)^2}{\left(2+\frac{1}{x}\right)}}_{\to \frac{1}{2}}.$$
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