Limit $\lim_{x \to 1} \frac{\log{x}}{x-1}$ without L'Hôpital I was wondering if it's possible to identify this limit without using L'Hôpital's Rule:
$$\lim_{x \to 1} \frac{\log{x}}{x-1}$$
 A: In THIS ANSWER and THIS ONE I showed, without the use of calculus, that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1$$
Therefore, we can write for $x>1$
$$\frac1x \le \frac{\log(x)}{x-1}\le 1$$
and for $x<1$
$$1 \le \frac{\log(x)}{x-1}\le \frac1x$$
whereupon applying the squeeze theorem yields the result $1$.
A: To elaborate on copper.hat's comment, it does not require L'Hopital's rule to recognize that
$$\lim_{x\to1}{\log x\over x-1}=\lim_{x\to1}{\log x-\log1\over x-1}=f'(1)\quad\text{where }f(x)=\log x$$
This is simply the definition of the derivative.  If you also know that $f'(x)=1/x$, then you get
$$\lim_{x\to1}{\log x\over x-1}=1$$
It might be noted that L'Hopital's rule is often invoked for problems like this, but its invocation is purely a matter of convenience.
A: Rewrite this using $t = h-1$ as $$\lim_{t\to 0} \frac{\log(1+t) - 0}{t}.$$
Now try to recognize this as the derivative of a function you know.
The point of this exercise is almost certainly to recognize that this is the definition of the derivative.
A: It's simple:
$$\lim_{x \to 1} \frac{\log{x}}{x-1}$$
$$=\lim_{(x-1) \to 0} \frac{\log{[1+(x-1)]}}{(x-1)}$$
$$=\lim_{u \to 0} \frac{\log{(1+u)}}{u}$$
$$= \log'(0)$$
$$=1$$
The last step comes from this link.
A: If you're looking to avoid relying on knowing derivatives already, you can write $\frac{\log x}{x-1}=\frac{\int_1^x\frac{dt}{t}}{x-1}$, showing it to be the mean value of $\frac{1}{t}$ for $t\in[1,x]$. Take $x\to1$ to show that the limit is equal to $1$.
A: Hint
Decompose the logarithm into Taylor series around $1$.
A: $$\lim_{x \to 1} \frac{\log{x}}{x-1}\quad\iff\quad\lim_{x\to 0}\frac{\log(x+1)}{x}$$
Using Taylor expansion we get
$$\lim_{x\to 0}\frac{x-x^2/2+O(x^3)}{x}=\lim_{x\to 0}(1-x/2+O(x^2))=1$$
A: 
From the above image it is easy to observe that,
 $$\frac{x-1}{x}<\int^{x}_{1}t dt<x-1$$
Since Area of smaller rectangle< Area under curve< Area of larger rectangle,
$$\rightarrow \frac{1}{x}<\frac{log(x)}{x-1}<1$$
$lim_{x \rightarrow 1} \frac{1}{x}=1$
Hence by Sandwich Theorem $lim_{x \rightarrow 1} \frac{log(x)}{x-1}=1$
