Question

Show that the limit of the function $f\left(x\right)=\frac{{\log }_{e}x}{x-1}$ at $x=1$ exist as Right hand and Left hand Limit and its value is 1

Answer 1

$\begin{array}{c}Righthandlimit\\ \Rightarrow {lim}_{x\to 1^{+}}f\left(1^{+}\right)={lim}_{h\to 0}f\left(1+h\right)=\frac{{\log }_e^{\left(1+h\right)}}{\left(1+h\right)-1}={lim}_{h\to 0}\frac{\frac{1}{\left(1+h\right)}}{1}=\frac{1}{\left(1+h\right)}=1\\ Lefthandlimit\\ \Rightarrow {lim}_{x\to 1^{-}}f\left(1^{-}\right)={lim}_{h\to 0}f\left(1-h\right){lim}_{h\to 0}\frac{{\log }_e^{\left(1-h\right)}}{\left(1-h\right)-1}=\frac{\frac{-1}{\left(1-h\right)}}{-1}={lim}_{h\to 0}\frac{1}{\left(1-h\right)}=1\\ \therefore f\left(1^{+}\right)=f\left(1^{-}\right)=1Ans.\end{array}$