# Finding a limit

#### SyNtHeSiS

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

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

MHF Hall of Fame
$$\displaystyle \lim_{x\to 1}\left(\frac{x}{x - 1} - \frac{1}{\ln x}\right)$$.

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

MHF Hall of Honor
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 $$\displaystyle \frac{1}{ \sqrt{x} - 1} = \frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1}$$ and put $$\displaystyle \ln{x} = 4\ln{\sqrt{x}}$$ 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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• Chris L T521

#### 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(Surprised)

#### drumist

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

• HallsofIvy

#### HallsofIvy

MHF Helper
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.