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I am confronted with the following limit and not sure how to argue rigorously: $$ \lim_{x \to \infty} 2x - \sqrt{4x^2+1} $$ Maple and Wolfram Alpha tell me its 0. Actually proving this is not hard either, since $2x - \sqrt{4x^2+1}$ is less than 0 for $x \geq 0$, increasing, and for any $\epsilon > 0$ there is an $x > 0$ such that $0 > 2x - \sqrt{4x^2+1} > -\epsilon$. Therefore it converges.

I am wondering if there is a faster way. I thought about something like $\sqrt{4x^2+1} \underset{x \to \infty}{\sim} 2x$ and therefore the limit must be 0. But you could argue $\sqrt{4x^2+1} \underset{x \to \infty}{\sim} 2x + c$ for any $c>0$ which means that any limit is possible, which is wrong.

So, is there a way one could immediately see that the limit must be 0?

julian2000P
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2 Answers2

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Multiplie by $\frac{2x+\sqrt{4x^2+1}}{2x+\sqrt{4x^2+1}}$ $$(2x-\sqrt{4x^2+1})\frac{2x+\sqrt{4x^2+1}}{2x+\sqrt{4x^2+1}}=\frac{4x^2-4x^2-1}{2x+\sqrt{4x^2+1}}=\frac{-1}{2x+\sqrt{4x^2+1}}$$ That clearly goes to zero as $x$ goes to $\infty$

Guillermo García Sáez
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If $x$ is positive $4x^2+1$ is trivially greater than $(2x)^2$ and less than $\left(2x+\frac{1}{4x}\right)^2$, so the limit as $x\to +\infty$ is zero by squeezing.

Jack D'Aurizio
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