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Given Limit: $L=\lim_{x\to 1} \frac{x}{\ln x}-\frac{1}{x \ln x}$
The limit reduces to $L=2$ via L'Hospital Rule
Even after employing the logarithmic limit $\lim_{x\to 0}\frac{\ln (1+x)}{x}=1$
Also after logarithmic series expansion $\ln (1+x) =x-\frac{x^2}{x}+\frac{x^3}{3}-\frac{x^4}{4}+...$
Couldn't attain the answer Please explain possible algebraic manipulation or procedure to figure out the solution.

Techno Freak
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4 Answers4

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HINT:

Note that we have

$$\frac{x}{\log(x)}-\frac1{x\log(x)}=\frac{x+1}{x}\frac{x-1}{\log(x)}$$

and

$$\frac{x-1}{x}\le \log(x)\le x-1$$

which can be obtained using the limit definition of the exponential function along with Bernoulli's inequality as I showed in THIS ANSWER.

Mark Viola
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Hint: set $y+1=x$ then $L=\lim_\limits{y\to 0} \frac{y+1}{\ln(y+1)}-\frac{1}{(y+1) \ln(y+1)}=\lim_\limits{y\to 0} \frac{y}{\ln(y+1)}\cdot\frac{y+2}{y+1}$

Vasili
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    And how do you evaluate the limit $\lim_{y\to 0}\frac{y}{\log(y+1)}$ without L'Hospital's Rule?? – Mark Viola Nov 22 '22 at 16:00
  • @MarkViola With either the definition of the derivative of the logarithm, or the Taylor series of $\ln(1+y)$. – J.G. Nov 22 '22 at 16:10
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    @J.G. And both of those are tantamount to LHR. – Mark Viola Nov 22 '22 at 16:15
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    @MarkViola: I see your point. The OP indicated that this limit is known to him/her. – Vasili Nov 22 '22 at 16:28
  • @Vasili Ah, indeed. Good catch. – Mark Viola Nov 22 '22 at 16:36
  • @MarkViola I disagree. L'Hôpital's rule generalizes the derivative-definition trick with the ratio of two in general nonlinear functions, while the Taylor series argument only requires computing a Maclaurin-series reciprocal to zeroth order. – J.G. Nov 22 '22 at 16:43
  • @J.G. Those are tantamount, the same in kind, not equivalent. And how does one find the derivative of the logvfunction? – Mark Viola Nov 22 '22 at 16:57
  • @MarkViola Well, if it's tantamount but not equivalent (I don't think it's even that), we've solved the problem without LHR as requested. If you define $\exp x$ as the $y_0=1$ solution of $y'=y$, its inverse $\ln x$ is $\int_1^x\frac1tdt$ for $x>0$. That $\ln'x=\frac1x$ on this definition is the basis for said Taylor series. – J.G. Nov 22 '22 at 17:29
  • @J.G. Yes, integral calculus is one way to proceed. But that requires even a more advanced arsenal. The OP is likely seeking a way forward that avoids calculus. – Mark Viola Nov 22 '22 at 17:54
  • With whichever definition of $\ln$ is deemed calculus-free. – J.G. Nov 22 '22 at 17:57
  • @J.G. As in the link I referenced, one can begin with the limit definition of the exponential function and use Bernoulli's inequality to arrive at elementary bounds for the exponential function. Then, define the logarithm function as the inverse of the exponential function and arrive at elementary bounds for the logarithm function. All tools are pre-calculus. ;-) – Mark Viola Nov 25 '22 at 02:18
  • I saw that link. It's a very lengthy strategy. If you're going to define then manipulate specific limits, you could just as easily use derivatives. You certainly needn't interpret LHR bans so broadly. Your answer may be legal given the OP's tags, but so is this one. – J.G. Nov 25 '22 at 07:46
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One can split terms into addition by applying logarithm again:

$$L = \lim_{x \to 1}{\frac{\frac{x^2 - 1}{x}}{\ln x}} = \lim_{x \to 1}\frac{\big(1 + \frac{1}{x}\big) \cdot \big(1 - \frac{1}{x}\big)}{\ln x}$$

Taking natural log:

$$\ln{L} = \lim_{x \to 1}{\ln{\big(1 + \frac{1}{x}\big)}} + \lim_{x \to 1}{\ln\big[{\frac{\big(1 + \frac{1}{x}\big)}{\ln{x}}}}\big]$$

The rest is an exercise.

Dstarred
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    And just how does that exercise work? – Mark Viola Nov 22 '22 at 16:38
  • @Unexpected Confusion Maybe that's a really long route and I am not sure of finding the solution with this approach. It probably may not be of much help during time contrained examinations where such limit evaluation are mere part of the main problem. Although still I am delighted that you took time to provide your way towards the solution. – Techno Freak Nov 22 '22 at 16:45
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We have $${x\over \ln x}-{1\over x\ln x}={x^2-1\over x\ln x}={2\over x}{x^2-1\over \ln(x^2)-\ln 1}$$ The limit is equal ${2\over \ln'(1)}=2.$

Ryszard Szwarc
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