Limit at infinity of $\frac{1}{x^{n-1}}$? We have $\lim_{x\rightarrow\infty}\frac{1}{x^n}=0$

But why is \begin{align} \lim_{x\rightarrow\infty}\frac{1}{x^{n-1}}=\lim_{x\rightarrow\infty}\frac{1}{x^n x^{-1}} =\lim_{x\rightarrow\infty}\frac{1}{\frac{x^n}{x}}=\lim_{x\rightarrow\infty}\frac{x}{x^n} \end{align} indeterminate?

Suppose I have the absolute value, is the limit still indeterminate? \begin{align} \lim_{x\rightarrow\infty}\Bigg\lvert \frac{1}{x^{n-1}}\Bigg\rvert &amp;=\lim_{x\rightarrow\infty}\Bigg\lvert \frac{1}{x^n x^{-1}}\Bigg\rvert \\ &amp;=\lim_{x\rightarrow\infty}\Bigg\lvert \frac{1}{\frac{x^n}{x}}\Bigg\rvert =\lim_{x\rightarrow\infty}\Bigg\lvert \frac{x}{x^n}\Bigg\rvert \end{align}
Is your question about $\displaystyle\lim_{x\to +\infty} \frac x {x^n}$ being zero, or about $\displaystyle\frac \infty \infty$ being zero? The first one is true (for $n &gt; 1$), the other one is indeterminate. Because for instance $\displaystyle\lim_{x\to +\infty} \frac {x^{2n}} {x^n} = +\infty$ is also of the form $\displaystyle\frac \infty \infty$.

The point is that $\displaystyle\frac \infty \infty$ is just a notation for a type of limits and not an equality with the previous term.
Indeterminate simply means "we don't know". It can be any number or even infinity. For example

$$\lim_{x\to+\infty} \frac{ax+1}{x}=\lim_{x\to+\infty} a+\frac{1}{x}=a$$

$$\lim_{x\to+\infty} \frac{ax^2+1}{x}=\lim_{x\to+\infty} ax+\frac{1}{x}=+\infty$$

Note that all these limite are indeterminate limites of the forms $\frac{+\infty}{+\infty}$.