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

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