l'hospital's rule

\[\lim \limits_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)} \rightarrow \frac{0}{0} \text{ or } \pm \frac{\infty}{ \infty}\]

L’Hospital’s Rule can be applied when the resulting limit is an indeterminate form.

\[\lim \limits_{x \to a} \frac{f'(x)}{g'(x)}\]

Sometimes, I have to apply L’Hospital’s Rule more than once.

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Prove \(\log n \in o(n^{0.01})\)

\[\lim \limits_{x \to \infty} \frac{\log n}{n^{0.01}}\] \[\lim \limits_{x \to \infty} \frac{\log n}{n^{0.01}} \approx \frac{\log \infty}{\infty^{0.01}} \rightarrow \frac{\infty}{ \infty}\] \[\text{apply L'Hospital's Rule}\] \[\begin{align*} \frac{d}{dx} \log n &= \frac{1}{n}\\ \frac{d}{dx} n^{0.01} &= 0.01 \cdot n^{-0.99}\\ \lim \limits_{x \to \infty} \frac{f'(x)}{g'(x)} &= \frac{\frac{1}{n}}{\frac{0.01}{n^{0.99}}}\\ &= \frac{n^{0.99}}{0.01 \cdot n}\\ &= \frac{n^{0.99 - 1}}{0.01}\\ &= \frac{100}{n^{0.01}}\\ &= \frac{100}{\infty}\\ &= 0 \end{align*}\]

This implies that \(n^{0.01}\) grows faster than \(\log n\), which means \(\log n \in o(n^{0.01})\)