Originally Posted by

**eniuqvw**

I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity).

__First Method__: make $\displaystyle 0$ the RHS,

$\displaystyle \begin{aligned}

\frac{1}{x}&<4 \\

\frac{1}{x}-4&<0 \\

\frac{1-4x}{x}&<0.

\end{aligned}$

To get a negative quantity we must study two cases:

- If $\displaystyle 1-4x<0$ & $\displaystyle x>0.$ Intersect these conditions and we have $\displaystyle S_1=\left] \frac{1}{4},\infty \right[.$

- If $\displaystyle 1-4x>0$ & $\displaystyle x<0.$ Intersect these conditions and we have $\displaystyle S_2=\bigg]-\infty,0 \bigg[.$

Final solution we'll be given by $\displaystyle S=S_{1}\cup S_{2}=\bigg] -\infty ,0 \bigg[\cup \left] \frac{1}{4},\infty \right[.$

__Second Method__: we obviously discard $\displaystyle x=0,$ otherwise the initial inequality makes no sense. So we're gonna multiply both sides by $\displaystyle x^2,$ since this quantity is positive, the inequality stays as it's written. Following up on this we have:

$\displaystyle \begin{aligned}

\frac{1}{x}&<4 \\

x&<4x^{2} \\

x(4x-1)&>0.

\end{aligned}$

From here study two cases again:

- If $\displaystyle 4x-1>0$ & $\displaystyle x>0\implies S_{1}=\left] \frac{1}{4},\infty \right[.$

- If $\displaystyle 4x-1<0$ & $\displaystyle x<0\implies S_{2}=\bigg] -\infty ,0 \bigg[.$

Hence $\displaystyle S=S_{1}\cup S_{2}=\bigg] -\infty ,0 \bigg[\cup \left] \frac{1}{4},\infty \right[,$ which is we got before.