Finding a limit

Apr 2010
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\(\displaystyle \lim_{x\to 1}\left(\frac{x}{x - 1} - \frac{1}{\ln x}\right)\).

I worked out that it was an interminate difference of the form ∞ - ∞, so I figured I have to differentiate xlnx - x + 1 / (x-1)(lnx), but then the answer will be reaaally long, and I dont know what to do. This sometimes happens when I try to differentiate something(Worried)
 
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Chris L T521

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\(\displaystyle \lim_{x\to 1}\left(\frac{x}{x - 1} - \frac{1}{\ln x}\right)\).

I worked out that it was an interminate difference of the form ∞ - ∞, so I figured I have to differentiate xlnx - x + 1 / (x-1)(lnx), but then the answer will be reaaally long, and I dont know what to do. This sometimes happens when I try to differentiate something(Worried)
That is almost the correct way of going about it!

\(\displaystyle \lim_{x\to1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)=\lim_{x\to 1}\frac{x\ln x - x+1}{(x-1)\ln x}\)

When you evaluate this limit, we get the indeterminate case \(\displaystyle \frac{0}{0}\). So to apply L'Hôpital's rule, we need to differentiate both numerator and denominator separately! Recall that L'Hôpital's rule says that if \(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}\) is of the form \(\displaystyle \frac{\infty}{\infty}\) or \(\displaystyle \frac{0}{0}\) and if \(\displaystyle \lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\) exists, then \(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\).

So we don't look at \(\displaystyle \lim_{x\to 1}\frac{d}{dx}\left[\frac{x\ln x-x+1}{(x-1)\ln x}\right]\), we look at \(\displaystyle \lim_{x\to1}\frac{\dfrac{d}{dx}[x\ln x-x+1]}{\dfrac{d}{dx}[(x-1)\ln x]}\)!

Does this clarify things? Can you finish this problem?
 

simplependulum

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Jan 2009
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I have an interesting method but not a normal ...


\(\displaystyle \frac{x}{x-1} - \frac{1}{\ln{x}} \)

\(\displaystyle = \frac{ (x-1) + 1}{x-1} - \frac{1}{\ln{x}} \)

\(\displaystyle = 1 + \left( \frac{1}{x-1} - \frac{1}{\ln{x}} \right )\)


I let \(\displaystyle L = \lim_{x\to 1} \left( \frac{1}{x-1} - \frac{1}{\ln{x}} \right ) \)

we know \(\displaystyle \frac{1}{t^2 - 1} = \frac{1}{2} \frac{1}{t-1} - \frac{1}{2} \frac{1}{t+1} \) and so

\(\displaystyle \frac{1}{x -1} = \frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1} \)

so we have

\(\displaystyle L = \lim_{x\to 1} \left(\frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1} - \frac{1}{\ln{x}} \right ) \)

also \(\displaystyle \ln{x} = 2\ln{\sqrt{x}}\)

\(\displaystyle L = \lim_{x\to 1} \left( \frac{1}{2} ( \frac{1}{\sqrt{x} -1} - \frac{1}{ \ln{\sqrt{x}}}) \right) - \frac{1}{2} \lim_{x\to 1} \frac{1}{\sqrt{x} + 1}\)

Let \(\displaystyle u = \sqrt{x} \)

\(\displaystyle L = \frac{L}{2} - \frac{1}{4} \)

\(\displaystyle L = -\frac{1}{2}\)

so the final answer is \(\displaystyle 1 - \frac{1}{2} = \frac{1}{2} \)


More surprisingly , if we continue to 'break down' the fraction \(\displaystyle \frac{1}{ \sqrt{x} - 1} = \frac{1}{2} \frac{1}{\sqrt[4]{x} -1} - \frac{1}{2} \frac{1}{\sqrt[4]{x}+1} \) and put \(\displaystyle \ln{x} = 4\ln{\sqrt[4]{x}}\) and so on for \(\displaystyle 8,16,32 ,...\) . All of the situations can reach the answer obtained !!

If there is no mistake in my method , it is interesting that the answer will not be affected for different choices of the base of the log. function !!

EDIT:

The answer will not be affected for different choices of the base of the log. function .
This statement is false because at first we don't know if the limit exists , we have \(\displaystyle L = \frac{L}{2} - \frac{1}{4} \) , but we are also forced to believe that \(\displaystyle \infty = \frac{ \infty}{2} - \frac{1}{4}\) . The limit only exists when the base is \(\displaystyle e \) .
 
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Apr 2010
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I tried differentiating and got:

lim x-> 1 ( lnx - 1 ) / ( (x - 1) / x ) + lnx ) and this ended up being ∞, yet the answer is 1/2(Surprised)
 
Jan 2010
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I tried differentiating and got:

lim x-> 1 ( lnx - 1 ) / ( (x - 1) / x ) + lnx ) and this ended up being ∞, yet the answer is 1/2(Surprised)
\(\displaystyle \frac{d}{dx} \left[ x \ln x - x + 1 \right] = \frac{x}{x} + \ln x - 1 = \ln x\)

\(\displaystyle \frac{d}{dx} \left[ (x-1) \ln x \right] = \frac{x-1}{x} + \ln x = 1 - \frac{1}{x} + \ln x\)

So: \(\displaystyle \lim_{x \to 1} \, \frac{\ln x}{1 - \frac{1}{x} + \ln x}\)

Again we get \(\displaystyle \frac{0}{0}\), so we repeat the process.

\(\displaystyle \frac{d}{dx} [\ln x] = \frac{1}{x}\)

\(\displaystyle \frac{d}{dx} \left[ 1 - \frac{1}{x} + \ln x \right] = \frac{1}{x^2} + \frac{1}{x}\)

So: \(\displaystyle \lim_{x \to 1} \, \frac{\frac{1}{x}}{\frac{1}{x^2} + \frac{1}{x}}\)
 
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HallsofIvy

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Apr 2005
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I tried differentiating and got:

lim x-> 1 ( lnx - 1 ) / ( (x - 1) / x ) + lnx ) and this ended up being ∞, yet the answer is 1/2(Surprised)
No, that is NOT \(\displaystyle \infty\), it is "\(\displaystyle \frac{0}{0}\)" so use L'Hopital's rule again.