Question

$\underset{x\to \epsilon }{lim}\frac{Inx-1}{|x-e|}\text{is equal to}$

• A. 1/e
• B. -1/e
• C. e
• D. Does not exist

Answer 1

The correct option is D

Does not exist

$\underset{h\to 0}{lim}\frac{In(e-h)-1}{e-h-e}\underset{h\to 0}{lim}\frac{lo{g}_{\epsilon }(1-\frac{h}{e})-1}{|-h|}L.H.L=\underset{h\to 0}{lim}\frac{loge+log(1-\frac{h}{e})-1}{h}\underset{h\to 0}{lim}\frac{-\frac{h}{e}-\frac{h^2}{2e^2}}{|le+h-e|}=-\frac{1}{e}=\underset{h\to 0}{lim}\frac{In(e+h)-1}{|e+h-e|}=\underset{h\to 0}{lim}\frac{loge(1+\frac{h}{e})}{|h|}=\underset{h\to 0}{lim}\frac{loge+log(1+\frac{h}{e})-1}{h}=\underset{h\to 0}{lim}\frac{\frac{h}{e}-\frac{h^2}{2e}}{h}=\frac{1}{e}\text{Since L.H.L}\ne \text{R.H.L}\therefore \underset{x\to \epsilon }{lim}\frac{Inx-1}{|x-e|}\text{does not exist}$