Logarithm function does not have discontinuity

Question

I cannot understand why the function $\displaystyle{\frac{\ln x}{x-1}}$ does not have a discontinuity at $x=1$?

Answer 1

At $x=1$ you have $\frac{0}{0}$, which is indeterminate. If you work out the limit of that function as $x$ approaches $1$, then you get (using L'Hopital's Rule):

$$\lim_{x\rightarrow 1} \frac{\ln x}{x-1}=\lim_{x\rightarrow 1}\frac{\frac{1}{x}}{1}=1.$$

Answer 2

Notice, let $x-1=t\implies t\to 0\ as\ x\to 1$ hence, $$\lim_{x\to 1}\frac{\ln x}{x-1}=\lim_{t\to 0}\frac{\ln (1+t)}{t}$$ Using Taylor's series expansion $\ln(t+1)$ $$=\lim_{t\to 0}\frac{t-\frac{t^2}{2}-\frac{t^3}{3}+O(t^3)}{t}$$ $$=\lim_{t\to 0}\left(1-\frac{t}{2}-\frac{t^2}{3}+O(t^2)\right)$$ $$=\left(1-0\right)=\color{red}{1}$$