Limit of the sequence $\lim_{n \to \infty}{\frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}}$

Question

Is there a way mathematically to prove this limit without using de l'hopital because I think I can't use it on sequences. I know I can say that the log function goes to infinity slower than the denominator but is that the only way I can prove this?

$$\lim_{n \to \infty}{\frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}=0}$$

Answer 1

Let $n=\mathrm e^m$. Note that as $n\to\infty$, we also have $m\to\infty$. Hence, observing that $0 \leq \frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}} \leq \frac{\log(n^3)}{\sqrt{n}}$, $$0\le\lim_{n \to \infty}{\frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}}\le\lim_{n \to \infty}3\frac{\log n}{\sqrt{n}}=\lim_{m \to \infty}3\frac{m}{\mathrm e^{m/2}}\le\lim_{m \to \infty}3\frac{m}{m^2/8}=0$$

observing that $\mathrm e^{x}>x^2/2,\, \forall x>0$.

Answer 2

Multiply $\sqrt{n}$ on the top and bottom: $$\lim_{n\to\infty}\frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}=\lim_{n\to\infty}\frac{\sqrt{n}\log(n^3-7)}{n+2}=\lim_{n\rightarrow\infty}\frac{\log\left((n^3-7)^\sqrt{n}\right)}{n+2}$$Since $\log n$ doesn't approach $\infty$ as fast as $n$, this becomes zero.

Edit: As a note, don't try comparing derivatives between the numerators. That would take way too much time and effort.

Answer 3

Note that we have for $n\ge2$

$$ \left|\frac{\log(n^3-7)}{\sqrt n+2/\sqrt n} \right|\le 3\frac{\log(n)}{\sqrt{n}}$$

Can you finish now?

Answer 4

Let $a_n = \frac{\log(n^3-7)}{\sqrt{n}+2/\sqrt{n}}$. Then $$0 \leq a_n \leq \frac{\log(n^3)}{\sqrt{n}} = 3\frac{\log(n)}{\sqrt{n}}.$$ So we may instead show that $\lim_{n\to \infty} b_n=0$ where $$b_n = \frac{\log(n)}{\sqrt{n}}.$$ To show that $\lim_{n\to \infty} b_n=0$ note that for any $0 < p < 1$ we can bound $$\log(n) = \int_1^n \frac{1}{x} \, dx \leq \int_1^n \frac{1}{x^p}\, dx = \frac{1}{1-p} (n^{1-p} - 1) \leq \frac{1}{1-p}n^{1-p}.$$ Setting, for example, $p=\frac34$, we have $$\log(n) \leq 4 \sqrt[4]{n},$$ and so $$0\leq b_n = \frac{\log(n)}{\sqrt{n}} \leq \frac{4}{\sqrt[4]{n}},$$ from which it follows that $\lim_{n\to \infty} b_n=0$.

Answer 5

Another way to show that ${\log n\over \sqrt{n}}$ tends to $0,$ in order to finish the answer by @Mark Viola, could be as follows. Let $16^k\le n< 16^{k+1}.$ Then $$\log n<(k+1)\log 16,\quad \sqrt{n}\ge 4^k$$ Hence $${\log n\over \sqrt{n}}\le \log 16\,{k+1\over 4^k}\le {4\log 2 \over 2^k} {k+1\over 2^k}\le {8\log 2 \over 2^{k+1}}\le {8\log 2 \over \sqrt[4]{n}}$$