$\displaystyle \lim_{x\to 0^+} \frac{ln x}{x}$
My answer: $\displaystyle \infty$
Correct answer: $\displaystyle -\infty$
And yes its $\displaystyle \lim_{x\to 0^+} NOT \lim_{x\to 0^-}$
I don't get it.
Graph y = ln(x)
Your insistence that it approaches from the positive side is unnecessary as the negative side is not in the Domain of the Real-valued logarithm function.
For x < 1, ln(x) < 0 isn't it?
I suspect your difficulty lies in the misapplication of the rule. Is ln(x)/x in indeterminate form around x = 0?
Ah I see why i can't use L'Hospital's rule (My book says L'Hospital with and S. without an S its L'Hôpital).
So this is what I get by looking at the graphs
$\displaystyle \lim_{x\to 0^+} \frac{ln x}{x}$
=$\displaystyle \frac{\lim_{x\to 0^+ }ln x}{\lim_{x\to 0^+ }x}$
= $\displaystyle \frac{-\infty}{0}$ = undefined
The answer in the back of the book is: $\displaystyle -\infty$
i'm sure that $\displaystyle \frac{-\infty}{0}$ = undefined
Either I'm doing something wrong and I don't see it or the back of the book is wrong.
Think of the limit as a really large neg number divide by a really small pos number.
Dividing by a really small pos number (eg 1/1000000) is the same as multiplying by a very large pos number eg 1000000. So you have a large neg x large pos to give a (really) large neg ie neg infinity.
Both versions are correct (though I personally prefer the spelling sans 's'.) Guillaume de l'Hôpital's name was originally spelled with an 's', but the French language has evolved and the 's' was dropped in favor of a circumflex accent. In French the use of a circumflex accent is often used to indicate the disappearance of a silent 's'. For example, the words tête (head), fenêtre (window), pâte (pasta), coût (cost), and of course hôpital (hospital) (compare with the Italian testa, finestra, pasta, costo, and ospedale) all use the circumflex in this way.
But his rule is universal, regardless of how we spell it.
Sometimes it could help if you use substitution:
$\displaystyle \lim_{x \to 0} x = \lim_{n \to \infty}\left(\frac1n \right)$
Your limit becomes:
$\displaystyle \lim_{x\to 0^+} \frac{ln x}{x} = \frac{\lim_{x\to 0^+ }(ln( x))}{\lim_{x\to 0^+ }(x)} = \frac{\lim_{n\to \infty }\left(ln\left(\frac1n\right)\right)}{\lim_{n\to \infty}\left(\frac1n \right)}$ = $\displaystyle \lim_{n\to \infty}(n \cdot (-\ln(x))$
The product is negative. The absolute values of both factors approach infinity. Therefore the complete product approaches negative infinity.
Compare y = ln(x) with y = ln(x)/x for x < 1.
You should be familiar with y = ln(x). It heads off down the negative y-axis as x approaches zero. Keeping that in mind, if we divide that be some positive number less than 1, our friend goes even faster down the negative y-axis.
You can show it.
f(x) = ln(x) ==> df/dx = 1/x Using x = 1/2 gives 2
Let's see if the other one is faster?
g(x) = ln(x)/x ==> dg/dx = [1-ln(x)]/(x^2) Using x = 1/2 gives 4(1+ln(2)) = 6.773. Recall that we are moving in the negative direction so a greater positive slope means a greater decrease.
Don't panic. Prove.
Don't guess. Show.
Don't assume. Verify.