# Find this limit without using L'Hôpitale

Can any give me a hint to solve this limit
$\lim_{x\to 1} \frac{1}{x-1}-\frac{1}{ln(x)}$
I used L'Hôpitale rule and I found that the solution is -$\frac{1}{2}$ And I tried another ways like
$T=x-1$
$\lim_{t\to 0} \frac{ln(t+1)-t}{tln(t+1)}$
And I don't get any thing after that
Any help would be very good
Note: taylor series isn't available Thanks so much :)

Let $t=e^x-1$ then you want to find (I changed the order here, so I get the limit $\frac{1}{2}$ instead of $-\frac{1}{2}$, sorry)

$$L=\lim\limits_{x\to 0} \frac{e^x-x-1}{x(e^x-1)}$$

multiply by $$1=\lim\limits_{x\to 0} \frac{e^x-1}{x}$$ and the problem simplifies to

$$L=\lim\limits_{x\to 0} \frac{e^x-x-1}{x^2}$$ Note that $$L=\lim\limits_{x\to 0} \frac{e^{-x}+x-1}{x^2}$$ thus

$$2L=\lim\limits_{x\to 0} \frac{e^{-x}+e^x-2}{x^2}= \lim\limits_{x\to 0} \left(\frac{1-e^{x}}{x}\right)\left(\frac{e^{-x}-1}{x}\right)=(-1)(-1)=1$$

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

• This (nice trick) assumes that the limit exists. – Did Oct 26 '16 at 18:12
• @did yeah, I was just waiting for somebody to ask that (they always do), the function is increasing as can easily be shown by algebra. – Rene Schipperus Oct 26 '16 at 18:19
• I see. So, the sign of $$\frac1{x(\log x)^2}-\frac1{x-1)^2}$$ is so obvious that no explanation about it is needed? :-) Not to mention that the function could be increasing on $x<1$ and on $x>1$ but have a jump at $1$... All in all, and trying to guess the OP's level of mathematical sophistication, I think I squarely disagree with your pedagogical choices here. – Did Oct 26 '16 at 18:30
• @did I dont care a jot about pedagogy. Why do you ? We are doing math here. – Rene Schipperus Oct 26 '16 at 18:32
• Interesting take. Anyway, where is the mathematical justification that the limits from the left and from the right exist and coincide? Without it, the answer is mathematically incomplete and falls into the most common mathematical trap about the subject. You want to "do maths"? Then do that properly! – Did Oct 26 '16 at 19:30