I have couple of question on this part of equation –
\begin{align}
\lim_{n\to \infty } \frac{ 7 \cdot \sqrt{n}}{\log(n)}- \lim_{n\to \infty} \frac{1}{n\cdot \log(n)} &=\lim_{n\to \infty} \frac{7 \cdot \sqrt{n}}{\log(n)}-0 \tag{1}\\
&=\lim_{n\to \infty}\frac{7 n}{2 \sqrt{n}} \tag{2}\\
&=\lim_{n\to \infty} \frac{7}{2} \sqrt{n} = \infty
\end{align}
1 – How does $$\lim_{n\to \infty} \frac{1}{n \log(n)}$$ evaluates to 0? Is it something to do with one of those limit rules that says limit x approaches to infinity 1/x^n = 0?
2 – How does log n in the equation evaluate to 2n^(1/2)?
Thanks
Best Answer
As it is hard to read the first part, i guess that it is the numbered stuff that is the question.
The first one as $\log(n)$ is greater 1 as $n>3$ we know that $$\frac{1}{n} \geq \frac{1}{\log(n) \cdot n} \geq 0$$
For the second questoin they use L'hospital, as $$\frac{d}{dx} \log(x)=\frac{1}{x}$$ and $$\frac{d}{dx} \sqrt{x}= \frac{1}{2\sqrt{x}}$$ we have $$7\cdot \frac{x}{2\sqrt{x}}= \frac{7}{2} \sqrt{x}$$