l'hospital rule

yoman360 · #1 $Old$ November 6th, 2009, 09:43 PM

$\lim_{x\to 0^+} \frac{ln x}{x}$

My answer: $\infty$

Correct answer: $-\infty$

And yes its $\lim_{x\to 0^+} NOT \lim_{x\to 0^-}$

I don't get it.

Debsta · #2 $Old$ November 6th, 2009, 09:58 PM

This is a case where L'Hopital's rule (by the way there's no S in Hopital) doesn't apply because you dont get 0/0 or inf/inf to begin with. The lim of the top is -inf, the lim of the bottom is 0, so you need another method.

TKHunny · #3 $Old$ November 6th, 2009, 10:03 PM

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?

yoman360 · #4 $Old$ November 6th, 2009, 10:39 PM

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
$\lim_{x\to 0^+} \frac{ln x}{x}$

= $\frac{\lim_{x\to 0^+ }ln x}{\lim_{x\to 0^+ }x}$

= $\frac{-\infty}{0}$ = undefined

The answer in the back of the book is: $-\infty$

i'm sure that $\frac{-\infty}{0}$ = undefined

Either I'm doing something wrong and I don't see it or the back of the book is wrong.

Debsta · #5 $Old$ November 6th, 2009, 11:06 PM

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.

Prove It · #6 $Old$ November 6th, 2009, 11:24 PM

Quote:

Originally Posted by Debsta $View Post$

This is a case where L'Hopital's rule (by the way there's no S in Hopital) doesn't apply because you dont get 0/0 or inf/inf to begin with. The lim of the top is -inf, the lim of the bottom is 0, so you need another method.

There IS an S in L'Hospital. It just happens to be a SILENT S.

Debsta · #7 $Old$ November 6th, 2009, 11:34 PM

Point taken. Please excuse my French.

redsoxfan325 · #8 $Old$ November 7th, 2009, 02:08 AM

Quote:

Originally Posted by Prove It $View Post$

There IS an S in L'Hospital. It just happens to be a SILENT S.

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.

earboth · #9 $Old$ November 7th, 2009, 02:35 AM

Quote:

Originally Posted by yoman360 $View Post$

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
$\lim_{x\to 0^+} \frac{ln x}{x}$

= $\frac{\lim_{x\to 0^+ }ln x}{\lim_{x\to 0^+ }x}$

...

Sometimes it could help if you use substitution:

$\lim_{x \to 0} x = \lim_{n \to \infty}\left(\frac1n \right)$

Your limit becomes:

$\lim_{x\to 0^+} \frac{ln x}{x} = \frac{\lim_{x\to 0^+ }(ln( x))}{\lim_{x\to 0^+ }(x)} = \frac{\lim_{n\to \infty }\left(ln\left(\frac1n\right)\right)}{\lim_{n\to \infty}\left(\frac1n \right)}$ = $\lim_{n\to \infty}(n \cdot (-\ln(x))$

The product is negative. The absolute values of both factors approach infinity. Therefore the complete product approaches negative infinity.

TKHunny · #10 $Old$ November 7th, 2009, 06:31 AM

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.