# Math Help - Finding a limit

1. ## Finding a limit

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

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

2. Originally Posted by SyNtHeSiS
$\lim_{x\to 1}\left(\frac{x}{x - 1} - \frac{1}{\ln x}\right)$.

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
That is almost the correct way of going about it!

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

When you evaluate this limit, we get the indeterminate case $\frac{0}{0}$. 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 $\lim_{x\to a}\frac{f(x)}{g(x)}$ is of the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$ and if $\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$.

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

Does this clarify things? Can you finish this problem?

3. I have an interesting method but not a normal ...

$\frac{x}{x-1} - \frac{1}{\ln{x}}$

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

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

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

we know $\frac{1}{t^2 - 1} = \frac{1}{2} \frac{1}{t-1} - \frac{1}{2} \frac{1}{t+1}$ and so

$\frac{1}{x -1} = \frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1}$

so we have

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

also $\ln{x} = 2\ln{\sqrt{x}}$

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

Let $u = \sqrt{x}$

$L = \frac{L}{2} - \frac{1}{4}$

$L = -\frac{1}{2}$

so the final answer is $1 - \frac{1}{2} = \frac{1}{2}$

More surprisingly , if we continue to 'break down' the fraction $\frac{1}{ \sqrt{x} - 1} = \frac{1}{2} \frac{1}{\sqrt[4]{x} -1} - \frac{1}{2} \frac{1}{\sqrt[4]{x}+1}$ and put $\ln{x} = 4\ln{\sqrt[4]{x}}$ and so on for $8,16,32 ,...$ . 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 $L = \frac{L}{2} - \frac{1}{4}$ , but we are also forced to believe that $\infty = \frac{ \infty}{2} - \frac{1}{4}$ . The limit only exists when the base is $e$ .

4. 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

5. Originally Posted by SyNtHeSiS
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
$\frac{d}{dx} \left[ x \ln x - x + 1 \right] = \frac{x}{x} + \ln x - 1 = \ln x$

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

So: $\lim_{x \to 1} \, \frac{\ln x}{1 - \frac{1}{x} + \ln x}$

Again we get $\frac{0}{0}$, so we repeat the process.

$\frac{d}{dx} [\ln x] = \frac{1}{x}$

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

So: $\lim_{x \to 1} \, \frac{\frac{1}{x}}{\frac{1}{x^2} + \frac{1}{x}}$

6. Originally Posted by SyNtHeSiS
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
No, that is NOT $\infty$, it is " $\frac{0}{0}$" so use L'Hopital's rule again.