What is the limit of the following function, as $x$ approaches $1$?

Solution 1:

You are correct in solving this problem by using L'Hôpital's rule. The answer -1 is also correct

Solution: $$\lim _{x\to 1}\left(\frac{ln\left(x\right)}{\arctan \left(1-x\right)}\right)$$ $$=\lim _{x\to 1}\left(\frac{\frac{d}{dx}\left(ln\left(x\right)\right)}{\frac{d}{dx}\left(\arctan \:\left(1-x\right)\right)}\right)$$ $$=\lim _{x\to 1}\left(\frac{\left(\frac{1}{x}\right)}{\left(\frac{-1}{\left(1-x\right)^2+1}\right)}\right)$$ $$=\lim _{x\to 1}\left(\frac{-x^2+2x-2}{x}\right)$$ $$=\left(\frac{-(1)^2+2(1)-2}{(1)}\right)$$ $$=\frac{-1}{1}$$ $$=-1$$

Solution 2:

$$ \lim_{x \to {1}} \frac{\ln(x)}{\arctan(1-x)}. $$ We must make the substitution here $x=(y+1)$, therefore $$ \lim_{y \to {0}} \frac{\ln(y+1)}{\arctan(-y)}=\lim_{y \to {0}} \frac{\ln(y+1)}{-\arctan(y)}. $$ I recommend to use here the Taylor series: $$ \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-....\tag{1} $$ $$ \arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-....,\quad\text{for }{-1\le{x}\le{1}},x\ne\pm{i}.\tag{2} $$ Therefore, $$ \lim_{y \to {0}} \frac{\ln(y+1)}{-\arctan(y)}=\lim_{y \to {0}} \frac{\left(y-\frac{y^2}{2}+\frac{y^3}{3}-....\right)}{-\left(y-\frac{y^3}{3}+\frac{y^5}{5}-....\right)}= \\ =\lim_{y \to {0}} \frac{y\left(1-\frac{y}{2}+\frac{y^2}{3}-....\right)}{-y\left(1-\frac{y^2}{3}+\frac{y^4}{5}-....\right)}=\lim_{y \to {0}}\frac{1}{-1}=-1. $$ We will have, that $$ \lim_{x \to {1}} \frac{\ln(x)}{\arctan(1-x)}=-1. $$