Given a curve $y = \frac{1}{t}$, for $t>0$, show that for $x>0$, $0<\frac {1}{3\ln(x)}<\sqrt[3]{x}$.

I know that $\int _1^{\sqrt[3]{x}}\frac{dt}{t}\:=\ln \:\sqrt[3]{x}$ but the width of the rectangle for $1<x<\sqrt[3]{x}$ is $$\sqrt[3]{x}-1$$

So how can $0<\frac {1}{3\ln(x)}<\sqrt[3]{x}$?

It should be $0<\frac {1}{3\ln(x)}<\sqrt[3]{x}-1$


We have $\sqrt[3]x-1 < \sqrt[3]x$.

If you have $0 < 1/3 \ln (x)< \sqrt[3]x-1$, then we must have

$$0 < 1/3 \ln (x)< \sqrt[3]x$$

  • $\begingroup$ I see. How studpid of me $\endgroup$
    – user71207
    Jul 20 at 11:12

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