How to evaluate $\lim_{x \to1}\frac{x- \ln x-1}{(x-1)^2}$ without L'Hopital?

Question

I will be thankful if somebody help me to solve this limit without L'Hospital's rule

$$\lim \limits_{x \to1}\frac{x- \ln x-1}{(x-1)^2}$$

Answer 1

HINT:

$$\log(x)=(x-1)-\frac12 (x-1)^2+O(x-1)^3$$

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Answer 2

If we use the change $u=x-1$,

then the limit becomes

$$\lim_{u\to 0}\frac{u-ln(1+u)}{u^2}$$

$$=\lim_{u\to 0}\frac{u-u+\frac{1}{2u^2}(1+\epsilon(u))}{u^2}$$

$$=\frac12.$$

Answer 3

Let $x=1+h.$ Then the expression equals

$$\tag 1 \frac{h - \ln (1+h)}{h^2} = \frac{1}{h^2}\int_1^{1+h}(1-1/t)\, dt.$$

Let $t= 1+s$ to see $(1)$ equals

$$\tag 2 \frac{1}{h^2}\int_0^{h}s/(1+s)\, ds.$$

Suppose $h>0.$ For small $s>0, 1-s < 1/(1+s) < 1-s +s^2,$ which gives $s-s^2 <s/(1+s) < s-s^2 + s^3.$ Thus $(2)$ is bounded below by $(h^2/2 -h^3/3)/h^2$ and bounded above by $(h^2/2 -h^3/3+h^4/4)/h^2.$ Both bounding expressions $\to 1/2$ as $h\to 0^+.$ Hence so does $(1).$ The argument is similar if $h\to 0^-.$

So, a longer proof, but we use no Taylor and no L'Hopital.