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$$\lim_{x \to 1} \frac{1+x}{\left(5x-3\right)\left(1+x^{2}\right)}=\frac{1}{2}$$

let $$f(x)=\frac{1+x}{\left(5x-3\right)\left(1+x^{2}\right)}$$ then $\forall \epsilon>0, \exists\delta>0,\forall x\in D_f \left(\left|x-\frac{1}{2}\right|<\delta\Longrightarrow \large\left|\frac{1+x}{\left(5x-3\right)\left(1+x^{2}\right)}-\frac{1}{2}\right|<\epsilon\right)$

$$\large\left|\frac{1+x}{\left(5x-3\right)\left(1+x^{2}\right)}-\frac{1}{2}\right|<\epsilon$$

$$\frac{\left|5x^{3}-3x^{2}+3x-5\right|}{\left|x^{2}+1\right|\left|5x-3\right|}<2\epsilon$$

but it seems that cannot be useful,so take $\delta\le1$ implies:$$\frac{-1}{2}<x<\frac{3}{2}$$$$\frac{1}{2}<-3x+5<\frac{13}{2}$$$$-5\left(\frac{3}{2}\right)^{3}<-5x^{3}<5\left(\frac{1}{2}\right)^{3}$$$$3\left(\frac{-1}{2}\right)^{2}<3x^{2}<3\left(\frac{3}{2}\right)^{2}$$

so:$$-5\left(\frac{3}{2}\right)^{3}+3\left(\frac{-1}{2}\right)^{2}+\frac{1}{2}<-5x^{3}+3x^{2}-3x+5<5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}$$ $$\left|-5x^{3}+3x^{2}-3x+5\right|<5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}$$$$\frac{\left|-5x^{3}+3x^{2}-3x+5\right|}{\left|x^{2}+1\right|}<\frac{5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}}{\left|x^{2}+1\right|}$$ since $$\frac{-11}{2}<\left|5x-3\right|<\frac{9}{2}$$we have:$$\frac{\left|-5x^{3}+3x^{2}-3x+5\right|}{\left|x^{2}+1\right|}<\frac{5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}}{\left|x^{2}+1\right|}<\frac{5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}}{\left|x^{2}+1\right|\frac{9}{2}}<\frac{\left(5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}\right)}{\left|x^{2}+1\right|\left|5x-3\right|}<2\epsilon$$, but this is wrong since $$\frac{5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}}{\left|x^{2}+1\right|}>\frac{5\left(\frac{1}{2}\right)^{3}+3\left(\frac{3}{2}\right)^{2}+\frac{13}{2}}{\left|x^{2}+1\right|\frac{9}{2}}$$, so how this kind of questions can be solved?

Hussain-Alqatari
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Absurd
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2 Answers2

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Let $f(x)=\dfrac{1+x}{(5x-3)(1+x^2)}.$ We want to show that $\forall \epsilon >0,\exists \delta >0$ st $|x-1|<\delta \Rightarrow \left|f(x)-\dfrac{1}{2}\right|<\epsilon.$

Simplify $f(x)-\dfrac{1}{2}$ as follows: $f(x)-\dfrac{1}{2}=\dfrac{1+x-\dfrac{1}{2}(5x-3)(1+x^2)}{(5x-3)(1+x^2)}$ $$\begin{align}=\dfrac{1+x-\dfrac{5}{2}x^3+\dfrac{3}{2}x^2-\dfrac{5}{2}x+\dfrac{3}{2}}{(5x-3)(1+x^2)}\end{align}$$ $$\begin{align}=\dfrac{-5x^3+3x^2-3x+5}{2(5x-3)(1+x^2)}\end{align}$$ $$\begin{align}=\dfrac{(x-1)(-5x^2-2x-5)}{2(5x-3)(1+x^2)}\end{align}$$.

Now let $\delta \leq \dfrac{1}{2}.$ Then $0.5<x<1.5$ and $\left|\dfrac{-5x^2-2x-5}{(5x-3)(1+x^2)}\right|<\left|\dfrac{19.25}{0.625}\right|=\dfrac{154}{5}.$ Thus, take $\delta =\min\{\dfrac{1}{2},\dfrac{5}{154}\epsilon\}.$ Then $|x-1|<\delta\Rightarrow\left|f(x)-\dfrac{1}{2}\right|<\dfrac{154|x-1|}{5}<\dfrac{154(\dfrac{5}{154}\epsilon)}{5}=\epsilon.$ Thus, by the $\epsilon-\delta$ definition, the limit is $\dfrac{1}{2}.$

Simon Fraser
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  • can you explain more where these $19.25$ and $0.625$ come from? – Absurd Oct 27 '19 at 07:05
  • If $x<1.5,$ then $|-5x^2-2x-5|<|5(1.5)^2+2(1.5)+5|=19.25.$ Similarly, if $x>0.5,$ then $|(5x-3)(1+x^2)|>|(5(0.5)-3)(1+0.5^2)|=0.625.$ Thus $\left|\dfrac{-5x^2-2x-5}{(5x-3)(1+x^2)}\right|<\left|\dfrac{-5x^2-2x-5}{0.625}\right|<\left|\dfrac{19.25}{0.625}\right|$ when $0.5<x<1.5.$ – Simon Fraser Oct 27 '19 at 07:10
  • Can I say $\frac{\left|x-1\right|\left|5x^{2}+2x+5\right|}{2\left|5x-3\right|\left|1+x^{2}\right|}<\frac{\left|x-1\right|\left|19.25\right|}{2\left|0.625\right|}<\frac{\left|19.25\right|\epsilon}{2\left|0.625\right|}=15.4\epsilon$ and then $\delta=\min\left{0.5,15.4\epsilon\right}$? – Absurd Oct 27 '19 at 07:46
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The problem is that the term $(5x-3)=0$ at $x=\frac35$.

Let wlog $|x-1|<\frac1{10}$ then we have

$$\left|\frac{1+x}{\left(5x-3\right)\left(1+x^{2}\right)}-\frac12\right|=\left|\frac{-5x^3+3x^2-3x+5}{2\left(5x-3\right)\left(1+x^{2}\right)}\right|=\left|\frac{-(x-1)(5x^2+2x+5)}{2\left(5x-3\right)\left(1+x^{2}\right)}\right|=\\=|x-1|\left|\frac{5x^2+2x+5}{2\left(5x-3\right)\left(1+x^{2}\right)}\right|<|x-1|\left|\frac{30}{2\left(1\right)\left(1\right)}\right|=15|x-1|$$

user
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  • can you please tell where does $30$ and $1$ come from? – Absurd Oct 28 '19 at 10:52
  • @Absurd since $0.9<x<1.1$ and $5x^2+2x+5$ is increasing for $x>0$ then take $x=2$ and $5x^2+2x+5=29<30$. it is a bound on the safe side. We are not forced to find an optimal bound. – user Oct 28 '19 at 10:57
  • I got it and what about $5x-3$ – Absurd Oct 28 '19 at 11:03
  • @Absurd also $5x-3 $ is increasing since it is linear and $5\cdot 0.9-3=\frac32>1$. – user Oct 28 '19 at 11:07
  • @Absurd The key point is to eliminate the point $x=\frac35$ from teh evaluation taking $|x-1|<1/10$ and then bound all term. – user Oct 28 '19 at 11:08