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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.

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    $\begingroup$ Try cancelling the x's $\endgroup$ – Matthew Levy Apr 2 '15 at 4:48
  • $\begingroup$ And what do you mean by "plugging infinity in"? $\endgroup$ – Prahlad Vaidyanathan Apr 2 '15 at 4:48
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    $\begingroup$ @MatthewLevy OMG THANK U SO MUCH, the answer is one! $\endgroup$ – Elsa Apr 2 '15 at 4:55
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$$\lim_{x\to \infty}\frac{\ln x}{\ln7x}$$

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

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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)!

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    $\begingroup$ I think you want a log on $7$. $\endgroup$ – Cameron Williams Apr 2 '15 at 5:03
  • $\begingroup$ Thanks to Cameron Williams, dh87 and any other editors for pointing this out! Typing too fast . . . Cheers! $\endgroup$ – Robert Lewis Apr 2 '15 at 5:10
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    $\begingroup$ I feel somewhat uncomfortable that this solution was not signed off with the usual "Fiat Lux!!!" $\endgroup$ – JessicaK Apr 2 '15 at 5:19
  • $\begingroup$ @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 math.stackexchange.com/questions/1090362/…, math.stackexchange.com/questions/1208587/… . . . continued $\endgroup$ – Robert Lewis Apr 2 '15 at 5:37
  • $\begingroup$ @JessicaK: (continuation) and the meta thread meta.math.stackexchange.com/questions/14999/… . . . in the meantime, thanks for your concern and, as ever, Fiat Lux!!! $\endgroup$ – Robert Lewis Apr 2 '15 at 5:39
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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.

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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.

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