l'hospitals rule help

**boomersooner1331** · November 15th 2009, 03:01 PM

i'm having a lot of trouble with these two problems.

the first one is tough, and i don't know how to do it, i just know the answer is supposed to be 1/2:

lim ((1/ln x) - (1/(1-x))
x --> 1

i found the derivatives of both and got:

((-1/(x ln x ^2) - ((1)/(x-1)^2)

but i don't know if that's what i'm supposed to do

and for the second one, i don't even know where to start:

lim (1 + (a/x)) ^ bx
x --> infinity

please....help. these are the last two of my homework problems and they're killing me!

**rubix** · November 15th 2009, 03:13 PM

remember, to apply L'Hospital's Rule you need

lim x --> c $f(x)/ g(x)$ where f(x) and g(x) are some function dependent on x. Furthermore, lim f(x) and lim g(x) both must be either 0 or +- infinity. This mear fact tell you that, you want to have a fraction of two function with x on numerator and denominator such that both have limit or either 0 or +- infinity.

once that is done, following holds:

lim x -> c $f(x)/g(x) =$ lim x -> c $f'(x)/g'(x)$

so what i'm trying to tell you here is this:

a) you want to have numerator and a denominator dependence on x. So, i would start by taking common denominator to get something of the form f(x) / g(x)

a) you want to make sure both lim f(x) and lim g(x) go to 0 or +- infinity before you use L'Hospital's Rule. So, taking derivative as you did is irrelevant when you clearly don't have a single numerator and a denominator and neither they go to 0 or +- infinity in given form.

**Veraltus** · November 15th 2009, 03:16 PM

Originally Posted by boomersooner1331

i'm having a lot of trouble with these two problems.

the first one is tough, and i don't know how to do it, i just know the answer is supposed to be 1/2:

lim ((1/ln x) - (1/(1-x))
x --> 1

i found the derivatives of both and got:

((-1/(x ln x ^2) - ((1)/(x-1)^2)

Well I know that you can only apply l'hopital's rule to an undeterminate form (0/0) or (infinity / infinity). I'm currently working on trying to solve it =D

**rubix** · November 15th 2009, 03:31 PM

i had enough time in my hand....

Originally Posted by boomersooner1331

lim ((1/ln x) - (1/(1-x))
x --> 1

$1/lnx - 1/ (1-x) = (1-x - lnx) / lnx (1-x)$

notice lim x --> 1 of top is 0 and bottom is 0 as well. Use L'Hospitals.

Originally Posted by boomersooner1331

lim (1 + (a/x)) ^ bx
x --> infinity

[math(1 + a/x) ^ (bx) = (x +a) ^ (bx) / (x)^bx

notice as x -> inf top and bottom goes to infinity. You know what to do next

P.S. i have not worked out the solution so idk how it works then forth...you might have to use L'hospitals rule once more.

**Veraltus** · November 15th 2009, 03:34 PM

Originally Posted by rubix

i had enough time in my hand....

$1/lnx - 1/ (1-x) = (1-x - lnx) / lnx (1-x)$

notice lim x --> 1 of top is 0 and bottom is 0 as well. Use L'Hospitals.

But if you go from common factor and use L'Hopital's rule you still get something that's not easily resolvable....

**boomersooner1331** · November 15th 2009, 03:36 PM

the second one is just one since they have the same powers and the coefficients are 1, right?

sorry if that seems like a dumb question. i'm absolutely exhausted. this has been the longest weekend of my life

**rubix** · November 15th 2009, 03:40 PM

Originally Posted by Veraltus

But if you go from common factor and use L'Hopital's rule you still get something that's not easily resolvable....

then use probably have to use L'Hospital's rule once more. In that case, you'll prolly need to use product rule on bottom.

**redsoxfan325** · November 15th 2009, 03:41 PM

Originally Posted by boomersooner1331

and for the second one, i don't even know where to start:

lim (1 + (a/x)) ^ bx
x --> infinity

please....help. these are the last two of my homework problems and they're killing me!

$(1+a/x)^{bx}=\exp\left[bx\ln(1+a/x)\right]$

Now you need to find the limit of the stuff inside the $\exp$ .

$\lim_{x\to\infty}\frac{b\ln(1+a/x)}{x^{-1}}=\lim_{x\to\infty}\frac{\frac{-ab}{x(x+a)}}{-x^{-2}}=$ $\lim_{x\to\infty}\frac{abx^2}{x(x+a)}=\lim_{x\to\i nfty}\frac{ab}{1+a/x^2}=ab$

So $\lim_{x\to\infty}(1+a/x)^{bx}=\lim_{x\to\infty}\exp\left[bx\ln(1+a/x)\right]=$ $\exp\left[\lim_{x\to\infty}bx\ln(1+a/x)\right]=\exp(ab)=\boxed{e^{ab}}$

**redsoxfan325** · November 16th 2009, 09:48 AM

Actually, I just thought of an easier way to do this:

$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{bx}=\ lim_{x\to\infty}\left[\left(1+\frac{1}{x/a}\right)^{x/a}\right]^{ab}=\left[\lim_{x\to\infty}\left(1+\frac{1}{x/a}\right)^{x/a}\right]^{ab}$

We know that $e$ is defined to be the stuff inside the brackets, so the final answer is $e^{ab}$ .

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