Question

ln2x / 7x limit ->0

Answer 1

do you do ln(2x) - ln(7x)?

Answer 2

Is it the ln(2x/7x) or the (ln(2x))/7x?

Answer 3

(ln(2x)) / 7x

Answer 4

I was going to suggest using L'Hopital's rule, but I noticed ln(0)=1 which does not help the entire thing come out to 0/0 as I hoped...Shucks.

Answer 5

Then no, you can't split it up as a subtraction only if it is inside the log.
\[\log(\frac{a}{b})=\log(a)-\log(b)\]
However, you have an indeterminant form, mainly: infty/0
So apply l'hospital's rule giving:
\[\lim_{x \rightarrow 0}\frac{\ln(2x)}{7x}=\lim_{x \rightarrow 0}\frac{\frac{d}{dx}\ln(2x)}{\frac{d}{dx}7x}=\lim_{x \rightarrow 0}\frac{\frac{2}{x}}{7}=\lim_{x \rightarrow 0}\frac{2}{7x} \rightarrow D.N.E.\]
The problem with this limit is (and I hate these) when you approach from one side you get -infty and the other you get +infty, so the limit technically does not exist.

Answer 6

You can see that here: http://www.wolframalpha.com/input/?i=limit+ln(2x)%2F7x%2Cx%2C0

Answer 7

You can see that here: http://www.wolframalpha.com/input/?i=limit+ln(2x)%2F7x%2Cx%2C0

Answer 8

@Tangent, the ln(1)=0; (ln(e^0)=0, since e^0=1) but ln(2)=undef