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I want to prove that $$\lim_{x\to\infty}\frac{x}{x^{2}+1} = 0.$$ So I start by saying, given $\varepsilon>0$ I want to find $M>0$ such that $$\forall x>M\implies\left|\frac{x}{x^{2}+1}-0\right|<\varepsilon.$$

Now everyone who has answered me so far has plucked an $M$ out of thin air without any indication as to how they have found it. So assuming we don't know $M$ how would we find it?

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  • $\begingroup$ Use $0 < \dfrac{x}{1+x^2} < \dfrac{1}{x}$, for $x > 0$ $\endgroup$ – r9m Dec 19 '14 at 0:28
  • $\begingroup$ It's often useful to look at the ratio of the dominant terms in the numerator and denominator. In this case, that ratio looks like 1/x, so you are led to consider proving it for 1/x and then adapting that to the original problem. $\endgroup$ – Neal Dec 19 '14 at 2:11
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The trick is to make the expression easier to deal with. For example, for $x > 0$

$$\left| \frac{x}{x^2 + 1} - 0 \right| \leq \left| \frac{x}{x^2} \right| = \frac{1}{x} $$

Now it becomes straight forward to find an appropriate $M$ given $\epsilon > 0$.

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  • $\begingroup$ But you have made something bigger than the original expression, surely we would want something smaller than it to ensure that what we have found is still less than $\varepsilon$? $\endgroup$ – user2850514 Dec 19 '14 at 0:30
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    $\begingroup$ Wrong way around. If you can bound $1/x$ by $\epsilon$, then the original expression is bound by $\epsilon$. $\endgroup$ – Simon S Dec 19 '14 at 0:31
  • $\begingroup$ Oh that makes complete sense, thank you. $\endgroup$ – user2850514 Dec 19 '14 at 0:32
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An important thing to note here is that this $M$ is not unique, you can take whatever $M$ that happens to work.

The most common way to do this is to "estimate" the thing somehow, making incremental changes to arrive at a simple expression.

$$\frac{x}{1+x^2} = \frac{1}{1/x+x} \le \frac{1}{x}$$ Thus if I can find an $M$ such that for $x > M$ I have $\frac{1}{x} < \epsilon$ then I'd be done. So I can choose $M = 1 / \epsilon $.

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