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I need to prove that $\lim_{x\to\infty} f(x)=1$ where $f(x)=(x^2+g(x))/(x^2+1)$ with $g:(0,\infty)\to\mathbb{R}$ and $|g(x)|\leq 5x$.

Here's my proof so far:

We need to prove that for all $\epsilon>0$ there exists an M>0 so that $|f(x)-1|<\epsilon$ for all x>M (this is the definition we use in class).

Thus, we have $|f(x)-1|=|(x^2+g(x))/(x^2+1)-1|=|(g(x)-1)/(x^2+1)|$. Now, since $x^2\geq 0$ for all x, $x^2+1>1$ and thus we get $|(g(x)-1)/(x^2+1)|=|g(x)-1|/(x^2+1)<|g(x)-1|/1=|g(x)-1|$

Here's where I get stuck. I know I need to somehow use the fact that $|g(x)|\leq 5x$ and after that find an M>0 such that for all x>M $|g(x)-1|<\epsilon$. Any help is appreciated!

pileafilea
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  • well, $(g(x) - 1)/(x^2 + 1) < (5x - 1)/(x^2 + 1)$. Now $\lim_{x \rightarrow +\infty} (5x -1)/(x^2 + 1) = 0$ Hence $f(x)$ will approach 1. The mistake in your reasoning is that you eliminate the square. – Ronald Oct 24 '22 at 13:41

1 Answers1

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By eliminating the square, you lost the driving force for the limit's behaviour.

Instead, you can do $$ 0\leq\Bigg|\frac{g(x)-1}{x^2+1}\Bigg|=\frac{|g(x)-1|}{x^2+1}\leq\frac{|5x|+1}{x^2+1}. $$ As the limit on the right goes to $0$ as $x\to\infty$, you are done. In terms of your $\varepsilon$, you need $$ \frac{5x+1}{x^2+1}<\varepsilon $$ (you can assume $x>0$ without loss of generality). So you are looking for those values of $x$ where $$ \varepsilon x^2-5x-(1-\varepsilon)>0. $$ The parabola will be positive to the right of its rightmost root. So you want $$ x>\frac{5+\sqrt{25+4\varepsilon(1-\varepsilon)}}{2\varepsilon} $$ to guarantee that $|f(x)-1|<\varepsilon$.

Martin Argerami
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