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The limit $$ \lim \limits_{n\to \infty }\left( \frac{15n+7}{n \log\left( n\right) } \right) $$ Is it valid to do

$\frac{15 + 7/n}{\log n}$

which makes the numerator go to $15$ as the bottom goes to infinity, therefore the whole thing goes to $0$? How would I show more formally that this is true?

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Your reasoning is perfectly fine, and can be made fully rigorous as follows. $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$


$\lim_{n\to\infty} \dfrac{15n+7}{n \ln(n)}$

$\ = \lim_{n\to\infty} \dfrac{15+7/n}{\ln(n)}$   [the division is valid because $n \ne 0$ and it does not change the value]

$\ = \dfrac{\lim_{n\to\infty}( 15+7/n )}{\lim_{n\to\infty} \ln(n)}$   [since these two limits exist in $\overline{\rr}$ and the ratio is defined in $\overline{\rr}$]

$\ = \dfrac{\lim_{n\to\infty} 15 + \lim_{n\to\infty} (7/n) }{\infty}$   [since these two limits exist in $\overline{\rr}$ and the sum is defined in $\overline{\rr}$]

$\ = \dfrac{15+0}{\infty} = 0$.


Note that this reasoning is carried out using the affinely extended reals $\overline{\rr}$, which includes $\infty$ and $-\infty$ but where certain operations are not allowed. The usual limit laws can be proven to work here as well as long as the extra restrictions as stated above are met.

Also, your question about validity of division by $n$ is a good one. Technically one has to specify the domain over which the limit is taken. I've never seen a textbook that specifies it, but it's important! For instance is $n$ a real number? Is it positive? In this case I assume you want $n \in \nn$ but you still need $n \ne 1$ otherwise $n \ln(n)$ would be zero. Personally I would write "$\lim_{n \in \nn_{>1} \to \infty} \dfrac{15n+7}{n \ln(n)}$" but I've never seen anyone else do that (they all rely on the readers to figure out the context...).

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  • $\begingroup$ @SeanHill: The method using the squeeze theorem does not capture the correct intuition that you had, so it is conceptually inferior to the method I've shown. However, the advantage is that it does not require working in $\overline{\rr}$, so you might see it in textbooks and normal analysis courses, which seem to avoid using $\overline{\rr}$ in general. $\endgroup$ – user21820 Aug 5 '16 at 15:13
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    $\begingroup$ +1 for the last paragraph which deals with issues which are rarely dealt in textbooks. $\endgroup$ – Paramanand Singh Aug 5 '16 at 18:31
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Holds the following $$0\leq\frac{15n + 7}{n\log n}\leq \frac{15n + 7n}{n\log{n}}\leq\frac{22}{\log{n}}<\varepsilon$$ for $n>e^\frac{22}{\varepsilon}$ and arbitrary $\varepsilon$.

By letting $\varepsilon\to 0$ and squeeze theorem, we can conclude $$\lim_{n\to\infty}\frac{15n + 7}{n\log n}=0.$$

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  • $\begingroup$ "for $n>e^\frac{22}{\varepsilon}$ and arbitrary $\varepsilon$" What? Where did this come from exactly? $\endgroup$ – Sean Hill Aug 5 '16 at 8:12
  • $\begingroup$ This came from the definition of limit. $\frac{22}{\log{n}}<\varepsilon$ is equivalent with $\log{n}>\frac{22}{\varepsilon}$, which is equivalent with $n>e^\frac{22}{\varepsilon}$. $\endgroup$ – alans Aug 5 '16 at 8:51
  • $\begingroup$ But as $\epsilon$ goes to $0$, $22/ \epsilon$ has denominator $0$ $\endgroup$ – Sean Hill Aug 5 '16 at 13:19
  • $\begingroup$ Yes, it means that $n$ goes to infinity. As $n$ becomes large and larger, $\frac{22}{n}$ becomes smaller and smaller and approaches $0$. $\endgroup$ – alans Aug 5 '16 at 14:45
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Your reasoning is fine but when you write it as an answer it needs to be made precise. Thus we have $$\lim_{n \to \infty}\frac{15n + 7}{n\log n} = \lim_{n \to \infty}\frac{15 + 7/n}{\log n} = \lim_{n \to \infty}(15 + 7/n)\cdot\frac{1}{\log n} = (15 + 0)\cdot 0 = 0$$ Often when a function $f(n) \to \infty$ as $n \to \infty$ then it is better to say that $1/f(n) \to 0$ as $n \to \infty$ and then we can use algebra of limits directly as I have shown above.

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Yes it is fully valid that $\lim_{n \to \infty} \frac{15n +7}{n \log n} = \lim_{n \to \infty} \frac{15 +\frac{7}{n}}{ \log n}$, and hence the limit is $0$.

To prove this, note that $\lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{1}{n}} =1$. So,

$\lim_{n \to \infty} \frac{15n +7}{n \log n} = \lim_{n \to \infty} \frac{\frac{1}{n}(15n +7)}{\frac{1}{n}(n \log n)} = \lim_{n \to \infty} \frac{15 +\frac{7}{n}}{ \log n} = 0$.

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  • $\begingroup$ I do not know why the answer earned a negative reputation point, $\endgroup$ – Dr. Dinesh J. Karia Feb 24 at 18:59
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$$\lim_{n\to \infty} \log n=\infty$$

That is enough to show that $$\lim_{n\to\infty}\frac{15n + 7}{n\log n}=0.$$

They are the key limits

$$\lim_{n\to \infty} \log n=\infty$$ $$\lim_{n\to \infty} \frac{n^{m_1}}{n^{m_2}}= \begin{cases} 0\quad, m_1<m_2\\ \infty\quad, m_1>m_2 \end{cases} $$ $$\lim_{n\to \infty} \frac{\log n}{n}=0$$

Always remember that the limit of logarithm is infinity but it has no chance to overcome polynomials.

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