finding limit of a given function using epsilon-delta definition

Question

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

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?

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Answer 1

The problem is that the term $(5x-3)=0$ at $x=\frac35$.

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

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

Answer 2

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 $\dfrac{1}{2}.$