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We have

$$\frac{n^b}{a^n}=e^{b\log n-n\log a}=e^{-n\left(\log a-b\frac{\log n}{n}\right)}\to 0$$

indeed

• $\log a-b\cdot \frac{\log n}{n} \to \log a-b\cdot 0=\log a>0$

and therefore

• ${n\cdot \left(\log a-b\cdot \frac{\log n}{n}\right)}\to \infty$

We have

$$\frac{n^b}{a^n}=e^{b\log n-n\log a}=e^{-n\left(\log a-b\frac{\log n}{n}\right)}\to 0$$

indeed

• $\log a-b\cdot \frac{\log n}{n} \to \log a-b\cdot 0=\log a>0$

and therefore

• ${n\cdot \left(\log a-b\cdot \frac{\log n}{n}\right)}\to \infty$

For $\frac{\log n}{n}\to 0$ refer to Prove $\log(x) \lt x^n$ for all $n \gt 0$.

$$\frac{n^b}{a^n}=e^{b\log n-n\log a}=e^{-n\left(\log a-b\frac{\log n}{n}\right)}\to e^{-\infty\cdot \left(\log a-b\cdot0\right)}=e^{-\infty}=0$$

We have

$$\frac{n^b}{a^n}=e^{b\log n-n\log a}=e^{-n\left(\log a-b\frac{\log n}{n}\right)}\to 0$$

indeed

• $\log a-b\cdot \frac{\log n}{n} \to \log a-b\cdot 0=\log a>0$

and therefore

• ${n\cdot \left(\log a-b\cdot \frac{\log n}{n}\right)}\to \infty$

$$\frac{n^b}{a^n}=e^{b\log n-n\log a}=e^{-n\left(\log a-b\frac{\log n}{n}\right)}\to e^{-\infty\left(\log a-b\cdot0\right)}=e^{-\infty}=0$$

$$\frac{n^b}{a^n}=e^{b\log n-n\log a}=e^{-n\left(\log a-b\frac{\log n}{n}\right)}\to e^{-\infty\cdot \left(\log a-b\cdot0\right)}=e^{-\infty}=0$$