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Let $f(x)=\frac{x^2+1}{x-3}$. I have to prove that $$\forall M>0, \exists \delta>0: |x-3|<\delta\implies f(x)>M$$

I proved that $$\frac{x^2+1}{x-3}=(x-3)+6+\frac{10}{x-3}$$ and thus, if $M>0$, $$f(x)>M\iff (x-3)(M-6)<10.$$

Here is my problem. If $M>6$, I just set $\delta=\frac{10}{M-6}$ to get $f(x)>M$. But if $M\in (0 ,6]$, which $\delta$ I can take ?

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  • $\begingroup$ Since you're taking the limit from the right, you should write $0\lt x-3\lt\delta$ in the first displayed expression, not $|x-3|\lt\delta$. $\endgroup$ – Barry Cipra Oct 31 '15 at 15:49
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Let $M>0$ be given. Set $\delta=\min \{ \frac{1}{M}, 1\}$. So if $|x-3|<\delta$ then \begin{equation} \frac{x^2+1}{x-3} \geq \frac{1}{x-3} > M . \end{equation}

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For $M\leq 6,$ just take, for instance, $\delta=10$. Then $f(x)>7,$ and is in particular greater than $M$. By the way, your if and only if does not hold- $f(x)>M$ if and only if $(x-3)(M-6)<10+(x-3)^2$, which does hold if (but not only if) $(x-3)(M-6)<10$. For instance when $x-3=10, f(x)=17$ is already much bigger than the corresponding $M=7$.

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  • $\begingroup$ Thanks for your answer. I would like to have something like $\delta=\min\{a,b\}$, does idm answer ($\delta=\min\{\frac{10}{|M-6|},1\}$) work ? I'm suspicions, because for $M=6$ it is not define. $\endgroup$ – Rick Oct 31 '15 at 11:25
  • $\begingroup$ Certainly it works. You just need to stipulate how your notation of min works when one piece is not defined. That said, there's no reason to tie yourself to the min form. $\endgroup$ – Kevin Carlson Oct 31 '15 at 14:57
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For $M<6$, any $\delta>0$ work. Then if you take for all $M>0$, $\delta=\min\left\{\frac{10}{|M-6|},1\right\}$ then $$f(x)>M$$

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  • $\begingroup$ Are you sure it work when $M=6$ ? $\endgroup$ – Rick Oct 31 '15 at 11:25
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Let $x>3$ and $M>0$.

$$\frac{x^2+1}{x-3}>M\iff x^2+1>M(x-3)$$ You know that $x^2+1>10$ for all $x>3$, then $$x^2+1>M(x-3)\iff M(x-3)<10\iff x-3<\frac{10}{M}$$

Just take $\delta=\frac{10}{M}$, and you'll have $$0<x-3<\delta\implies \frac{x^2+1}{x-3}>M.$$

You don't need other restriction... The only thing is if you would have that your function is defined on $(3,4]$ only. Then you would have the restriction that $3+\delta\leq 4$, and indeed, you can set $\delta=\min\{1,\frac{10}{M}\}$. But even in the case it's not necessary since ou want that $$\forall x\in (3,4], |x-3|<\delta\implies \frac{x^2+1}{x-3}>M$$ and thus, if $\delta=50$ you will naturally have that $x<\delta$ since $x<4$.

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Hint: I give you a proof of the statement, from which you should be able to solve your problem:

If $x > 3$, then $$ \frac{x^{2}+1}{x-3} > \frac{x}{x-3} > \frac{3}{x-3}; $$ if $M > 0$, then $ 0 < x-3 < 3/M $ only if $$ \frac{3}{x-3} > M. $$ We have proved this: for every $M > 0$, we have $0 < x-3 < 3/M$ only if $$ \frac{x^{2}+1}{x-3} > M. $$

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  • $\begingroup$ Can I also do like: Suppose $3<x<4$, then $$\frac{x^2+1}{x-3}>M\iff x-3<\frac{x^2+1}{M}\iff x-3<\frac{17}{M}$$ then, if I take $\delta=\min\{1,\frac{17}{M}\}$ we have that $0<x-3<\delta\implies \frac{x^2+1}{x-3}>M$. ? $\endgroup$ – Rick Oct 31 '15 at 13:48
  • $\begingroup$ No it doesn't work since if $\delta=1$ then $\frac{x^2+1}{x-3}>\frac{10}{17}M<M$. See my answer below. $\endgroup$ – Surb Oct 31 '15 at 14:18

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