L Hopital Rule for the limit (as x approaches 1) of lnx * ln(x-1)

[endrits079]

Im not sure if Im doing something wrong
but in this problem

lim xapproaches1 of lnx*ln(x-1) , to use the l hopital rule you have to turn it as a refractional and you can write it either as lnx/(1/ln(x-1)) or ln(x-1)/(1/lnx)
but the result is different can someone try if i did wrong something if not how comes it to have 2 different results

 

[Deleted member 4993]

Im not sure if Im doing something wrong
but in this problem

lim xapproaches1 of lnx*ln(x-1) , to use the l hopital rule you have to turn it as a refractional and you can write it either as lnx/(1/ln(x-1)) or ln(x-1)/(1/lnx)
but the result is different can someone try if i did wrong something if not how comes it to have 2 different results

We cannot tell "what you did wrong" - unless we see your work!
Please share your work with us ...even if you know it is wrong If you are stuck at the beginning tell us and we'll start with the definitions. You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL: http://www.freemathhelp.com/forum/th...Before-Posting

 

[endrits079]

here is it

this is my work

 

[ksdhart]

Generally speaking, if you try two (or more) different methods and get two different answers, it means at least one of them is wrong. In this case, your second method produces the actual value of the limit, and your first method is wrong. Specifically, it's this step here that's wrong:

$\displaystyle \displaystyle \lim _{x\to 1}\left(\frac{\left(x-1\right)\cdot ln^2\left(x-1\right)}{-x}\right)=\lim _{x\to 1}\left(\frac{\left(x-1\right)\cdot ln^2\left(x-1\right)+\left(x-1\right)\cdot ln^2\left(x-1\right)}{-1}\right)$

I'm guessing you meant to apply L'Hopital's Rule here, but you can't because as x goes to 1, the denominator is just -1. So you don't have one of the indeterminate forms required. Furthermore, even if could have used L'Hopital's rule, the derivative of the numerator is wrong too:

$\displaystyle \frac{d}{dx}\left(\left(x-1\right)\cdot ln^2\left(x-1\right)\right)\ne 2\cdot \left(x-1\right)\cdot ln^2\left(x-1\right)$

As you've seen, sometimes when one way gets ugly and complicated, it's better to just start over and try a different method which will hopefully be easier. Your first method will eventually lead to the correct answer, but it's not going to be pleasant.

$\displaystyle \displaystyle \lim _{x\to 1}\left(\frac{\left(x-1\right)\cdot ln^2\left(x-1\right)}{-x}\right)=\lim _{x\to 1}\left(\frac{x-1}{-\frac{x}{ln^2\left(x-1\right)}}\right)=\lim _{x\to 1}\left(\frac{1}{\frac{d}{dx}\left(-\frac{x}{ln^2\left(x-1\right)}\right)}\right)=\lim _{x\to 1}\left(\frac{1}{\frac{\frac{2x}{x-1}-ln\left(x-1\right)}{ln^3\left(x-1\right)}}\right)$

And on and on it goes...

 

[endrits079]

where did i make it wrong

could you tell me why my numerator derivative is wrong this is exactly what i did
d/dx(x-1)*ln^2(x-1)=d/dx(x-1)*ln^2(x-1)+(x-1)*d/dx ln^2(x-1) the derivative of (x-1) is just 1 so i wroge only
ln^2(x-1) and the derivative of ln^2(x-1) using chain rule is
d/dxln^2(x-1)*d/dxln(x-1)*d/dx(x-1)
where it is 2ln(x-1)*1/(x-1)*1 now x-1 with 1/x-1 simplifies and its 1 where it comes all this is ln^2(x-1)+2ln(x-1)

 

[ksdhart]

Oh, okay. I misunderstood what you meant in your handwritten work. Upon taking a closer look, I see that you used prime notation to indicate derivatives. I had interpreted these as showing an exponent of 1. So, yes, the derivative of the numerator was correct. Sorry about the confusion.

 

[stapel]

could you tell me why my numerator derivative is wrong this is exactly what i did
d/dx(x-1)*ln^2(x-1)=d/dx(x-1)*ln^2(x-1)+(x-1)*d/dx ln^2(x-1) the derivative of (x-1) is just 1 so i wroge only
ln^2(x-1) and the derivative of ln^2(x-1) using chain rule is
d/dxln^2(x-1)*d/dxln(x-1)*d/dx(x-1)
where it is 2ln(x-1)*1/(x-1)*1 now x-1 with 1/x-1 simplifies and its 1 where it comes all this is ln^2(x-1)+2ln(x-1)

The above is difficult to read. (That's not a criticism; some things simply don't type well. ) Is the following a correct interpretation?

. . . . .\(\displaystyle \begin{align}\dfrac{d}{dx}\, \bigg[\,(x\, -\, 1)\, \cdot\, \ln^2(x\, -\, 1)\,\bigg]\,

&=\, \dfrac{d}{dx}\, (x\, -\, 1)\, \cdot\, \ln^2(x\, -\, 1)\, +\, (x\, -\, 1)\, \dfrac{d}{dx}\, \ln^2(x\, -\, 1)

\\ \\ &=\, (1)\, \cdot\, \ln^2(x\, -\, 1)\, +\, (x\, -\, 1)\, \cdot\, \dfrac{d}{dx}\, \ln^2(x\, -\, 1)\, \cdot\, \dfrac{d}{dx}\, \ln(x\, -\, 1)\, \cdot \, \dfrac{d}{dx}\, (x\, -\, 1)

\\ \\ &=\, \ln^2(x\, -\, 1)\, +\, (x\, -\, 1)\, \cdot\, 2\ln(x\, -\, 1)\, \cdot\, \dfrac{1}{x\, -\, 1}\, \cdot 1

\\ \\ &=\, \ln^2(x\, -\, 1)\, +\, 2\ln(x\, -\, 1) \end{align}\)

By the way, Wolfram Alpha agrees (under "Alternate Form") with your derivative for (x - 1)*ln²(x - 1).

 

[endrits079]

The above is difficult to read. (That's not a criticism; some things simply don't type well. ) Is the following a correct interpretation?

. . . . .\(\displaystyle \begin{align}\dfrac{d}{dx}\, \bigg[\,(x\, -\, 1)\, \cdot\, \ln^2(x\, -\, 1)\,\bigg]\,

&=\, \dfrac{d}{dx}\, (x\, -\, 1)\, \cdot\, \ln^2(x\, -\, 1)\, +\, (x\, -\, 1)\, \dfrac{d}{dx}\, \ln^2(x\, -\, 1)

\\ \\ &=\, (1)\, \cdot\, \ln^2(x\, -\, 1)\, +\, (x\, -\, 1)\, \cdot\, \dfrac{d}{dx}\, \ln^2(x\, -\, 1)\, \cdot\, \dfrac{d}{dx}\, \ln(x\, -\, 1)\, \cdot \, \dfrac{d}{dx}\, (x\, -\, 1)

\\ \\ &=\, \ln^2(x\, -\, 1)\, +\, (x\, -\, 1)\, \cdot\, 2\ln(x\, -\, 1)\, \cdot\, \dfrac{1}{x\, -\, 1}\, \cdot 1

\\ \\ &=\, \ln^2(x\, -\, 1)\, +\, 2\ln(x\, -\, 1) \end{align}\)

By the way, Wolfram Alpha agrees (under "Alternate Form") with your derivative for (x - 1)*ln²(x - 1).

yes thats what i mean

 

[endrits079]

so have you come to a solution how is it possible to have two different results

 

[ksdhart]

so have you come to a solution how is it possible to have two different results

Yes. As I explained in my previous reply, you got two different answers because one of them is wrong. You tried to apply L'Hopital's Rule in a situation where the rule doesn't apply (because the denominator goes to -1). The second method you used lead to the correct answer of 0. The first one will as well, but it's much nastier to work with.

 

[endrits079]

actually it is -1 when i derivated it the derivative of -x is -1 what do you mean cant apply l hopitals rule

 

[pka]

what do you mean cant apply l hopitals rule

Have a look at this page. Scroll down to the $\displaystyle 0\cdot\infty$ form.

Can this function have a left-hand limit?

 

[endrits079]

Have a look at this page. Scroll down to the $\displaystyle 0\cdot\infty$ form.

Can this function have a left-hand limit?

what do you mean left hand limit i dont understand

 

[stapel]

what do you mean left hand limit i dont understand

Look in the index of your textbook (in the back of the book) for "limit". There should be sublistings for right- and left-hand limits. You'll see there that these refer, respectively, to the limits as one "approaches" the target x-value from the right-hand side (the "greater than" side of the number line) or the left-hand side. In your case, the left-hand limit, as x approaches 1, would be -3, -2, -1, 0, 0.5, 0.9, 0.99, 0.999, etc.

Is the log defined for these x-values?

 

[pka]

what do you mean left hand limit i dont understand

The function in the original post is $\displaystyle \large f(x)=\log(x)\cdot\log(x-1)$
That function is only defined if $\displaystyle x>1$.
Therefore can $\displaystyle \Large\displaystyle{\lim _{x \to {1^ - }}}f(x)$ exist?