$\lim\limits_{x\to\infty}\frac{\log x}{\log 7x}$?

Question

I did l'Hospital's rule and I got

$\dfrac{1}{x}/\dfrac{7}{7x} = \dfrac{0}{0}$ (after "plugging" $\infty$ in)

But is that the answer? I dont think $\frac{0}{0}$ is an answer. Or is it the answer and that means the limit is divergent.

And what do you mean by "plugging infinity in"? – Prahlad Vaidyanathan Apr 02 '15 at 04:48 — Prahlad Vaidyanathan, Apr 02 '15 at 04:48
@MatthewLevy OMG THANK U SO MUCH, the answer is one! – Elsa Apr 02 '15 at 04:55 — Elsa, Apr 02 '15 at 04:55

score 5 · Answer 1 · edited Apr 02 '15 at 05:09

5

To properly exploit L'hopital's rule, one must know the limit exists; see this wiki article. Here we kill two birds with one stone, showing that there is a limit and that it is $1$, without using L'hopital or derivatives:

For $x > 1$, $\ln x > 0$, whence we can say:

$\dfrac{\ln x}{\ln 7x} = \dfrac{\ln x}{\ln x + \ln7} = \dfrac{1}{1 + \dfrac{\ln7}{\ln x}} \to 1 \tag{1}$

as $\ln x \to \infty$. As $x \to \infty$, $\ln x \to \infty$, and so . . . voila!, followeth (1)!

edited Apr 02 '15 at 05:09

dh16

1,418

answered Apr 02 '15 at 04:59

Robert Lewis

71,180

3

I think you want a log on $7$. – Cameron Williams Apr 02 '15 at 05:03
Thanks to Cameron Williams, dh87 and any other editors for pointing this out! Typing too fast . . . Cheers! – Robert Lewis Apr 02 '15 at 05:10
1

I feel somewhat uncomfortable that this solution was not signed off with the usual "Fiat Lux!!!" – JessicaK Apr 02 '15 at 05:19
@JessicaK: Me too. *Fiat Lux!!!* etc. incurred the displeasure of at least one moderator and other "authorities", so I somewhat ruefully ceased using it. I am still corresponding with said "authorities" on this. For more, see the comment streams to http://math.stackexchange.com/questions/1090362/if-every-vector-is-an-eigenvector-the-operator-must-be-a-scalar-multiple-of-the, http://math.stackexchange.com/questions/1208587/if-r-is-a-pid-is-it-true-that-r-ker-phi-is-also-a-pid . . . continued – Robert Lewis Apr 02 '15 at 05:37
@JessicaK: (continuation) and the meta thread http://meta.math.stackexchange.com/questions/14999/how-to-deal-with-users-padding-their-answers-with-custom-signatures . . . in the meantime, thanks for your concern and, as ever, *Fiat Lux!!!* – Robert Lewis Apr 02 '15 at 05:39

score 2 · Accepted Answer · edited Apr 04 '15 at 20:30

2

$$\lim_{x\to \infty}\frac{\ln x}{\ln7x}$$

Let $f(x)=\dfrac{\ln x}{\ln7x}$ as you it goes to in determinant, so L'Hopital's rule gives $$\frac{\frac{1}{x}}{\frac{7}{7x}}=\frac{7x}{7x}=1$$

i.e., the limit value is 1

edited Apr 04 '15 at 20:30

Edward Jiang

3,670

answered Apr 02 '15 at 04:59

Quality

5,527

score 1 · Answer 3 · answered Apr 02 '15 at 06:04

Something else to note, using property of logs, we have $\log (7x)=\log(7) +\log (x)$. So your fraction can be rewritten as

$\frac {\log x}{\log x +\log 7}$. Since $\log 7$ is constant, as $x\to \infty$, we have $\log x \to \infty$, so the tiny increase is meaningless, thus a limit of 1. Formally, you'd still probably use L'Hospitals as above.

score 0 · Answer 4 · answered Apr 02 '15 at 07:00

Don't be hasty to put the x value too soon: $$\require{cancel}\lim_{x\to\infty}\frac{\ln x}{\ln 7x}\stackrel{\rm L'Hos.}\lim_{x\to\infty}\frac{\frac1{\cancel{\color{blue}x}}}{\frac{1}{\cancel{\color{red}7}\cancel{\color{blue}x}}.\cancel{\color{red}7}}=1$$ where I have used the chain rule for log7x.

$\lim\limits_{x\to\infty}\frac{\log x}{\log 7x}$?

4 Answers4