L' Hopital's Rule on a seriously rigged problem

**vstexas09** · October 24th 2008, 02:32 PM

In this exercise, (a) describe the type of indeterminate form (if any) that is obtained by direct substitution, (b) evaluate the limit using LHopital's rule if necessary, (C) use a graphing utility to graph the function and verify the result in part (B)

I tried so hard; and the solution online on "calcchat.com" doesn't make any sense...if you want to look it up, go to the website, and go to solutions, the book is calculus 8th edition, chapter 8, section 7, problem 49...

the solution makes no sens

here's the problem...

lim (as x approaches 1 from the right) = (ln x) ^ x-1

i tried...but i can't figure it out...

**galactus** · October 24th 2008, 03:51 PM

Is that ${\displaystyle}\lim_{x\to 1^{+}}(ln(x))^{x-1}$

or

$\lim_{x\to 1^{+}}(ln(x))^{x}-1$

I am guessing the former.

If so, it is the form $0^{0}$

Since the ol' hospital rule is the topic at hand, we can try this:

Rewrite:

$\lim_{x\to 1^{+}}e^{(x-1)ln(ln(x))}$

L'Hopital:

$e^{\displaystyle{-\lim_{x\to 1^{+}}\frac{(x-1)^{2}}{xln(x)}}}$

quotient rule:

${e^{\displaystyle{\left(\frac{-\lim_{x\to 1^{+}}\frac{(x-1)^{2}}{ln(x)}}{\lim_{x\to 1^{+}}x}\right)}}}$

The limit in the denominator of the above is 1, so we have:

$e^{\displaystyle{\left(-\lim_{x\to 1^{+}}\frac{(x-1)^{2}}{ln(x)}\right)}}$

L'Hopital again:

$e^{\displaystyle{\left(-2\lim_{x\to 1^{+}}x(x-1)\right)}}$

Now, we could continue using sum rules and product rules and what not, but it is easy to see what the limit is now. See it?.

**o_O** · October 24th 2008, 04:00 PM

Is that $\lim_{x\to 1^{+}}(\ln(x))^{x-1}$

or

$\lim_{x\to 1^{+}}(\ln (x))^{x}-1$

I am guessing the former.

If so, it is the form $0^{0}$

Since the ol' hospital rule is the topic at hand, we can try this:

Rewrite:

$\lim_{x\to 1^{+}}e^{ \displaystyle (x-1) \ln (\ln(x))}$

L'Hopital:

$e^{\displaystyle \left( -\lim_{x\to 1^{+}}\frac{(x-1)^{2}}{x \ln (x)}\right) }$

quotient rule:

${\Large}{e^{\left(\frac{\displaystyle -\lim_{x \to 1^{+}}\frac{(x-1)^{2}}{\ln (x)}}{\displaystyle \lim_{x\to 1^{+}}x}\right)}}$

The limit in the denominator of the above is 1, so we have:

$e^{\displaystyle \left(-\lim_{x\to 1^{+}}\frac{(x-1)^{2}}{\ln (x)}\right)}$

L'Hopital again:

$e^{\displaystyle \left(-2\lim_{x\to 1^{+}}x(x-1)\right)}$

Now, we could continue using sum rules and product rules and what not, but it is easy to see what the limit is now. See it?.

Does anyone know the trick to get the limit to display nicely instead of all spread out like like?. Also, how do we make the font larger?. I can not get the displaystyle to work on this site.

.

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