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: $$\lim_{x\to\infty}\dfrac{x^2+4-4x}{2x+1}=\lim_{x\to\infty}\dfrac{({x\over x}-{2\over x})({x\over x}-{2\over x})}{({2x\over x}+{1\over x})}=\dfrac{(1-0)(1-0)}{(2+0)}={1 \over 2}$$

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You have divided the numerator by $x^2$, but the denominator only by $x$. – Daniel Fischer Apr 7 '14 at 10:39
What do you mean? I divided both numerator and dneominator by x – Mouse Hello Apr 7 '14 at 10:40
No, $$(x-2)^2 = (x-2)(x-2) = \left[x\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\right]\cdot\left[x\left(\tfrac{x}{‌​x}-\tfrac{2}{x}\right)\right] = x^2\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\left(\tfrac{x}{x}-\tfrac{2}{x}\right).‌​$$ – Daniel Fischer Apr 7 '14 at 10:43
@DanielFischer are you going to write it as an answer? – Mouse Hello Apr 7 '14 at 10:45

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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The online calculator is correct. The value of the limit $\to\infty$. Here is the proof: \begin{align} \lim_{x\to\infty}{(x-2)^2\over2x+1}&=\lim_{x\to\infty}\frac{x^2-4x+4}{2x+1}\\ &=\lim_{x\to\infty}\frac{x^2-4x+4}{2x+1}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\\ &=\lim_{x\to\infty}\frac{\frac{x^2}{x}-\frac{4x}{x}+\frac{4}{x}}{\frac{2x}{x}+\frac{1}{x}}\\ &=\lim_{x\to\infty}\frac{x-4+\frac{4}{x}}{2+\frac{1}{x}}\\ &=\frac{\left(\lim\limits_{x\to\infty}x\right)-4+0}{2+0}\to\boxed{\Large\color{blue}{\infty}}\\ \end{align}

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$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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