Question

The value of the limit $\underset{x\to e}{\lim }\frac{\log x-1}{x-e}$ is

• A. $e$
• B. $\frac{1}{e}$
• C. $-\frac{1}{e}$
• D. $-e$

Answer 1

The correct option is B $\frac{1}{e}$

$\underset{x\to e}{\lim }\frac{\log x-1}{x-e}\mathrm{Let}x-e=h.\mathrm{Now},x=e+h\mathrm{if}x\to e\mathrm{then}h\to 0\mathrm{Therefore},\Rightarrow \underset{h\to 0}{\lim }\frac{\log \left(e+h\right)-\log e}{e+h-e}=\underset{h\to 0}{\lim }\frac{\log \left(1+\frac{h}{e}\right)}{h}=\underset{h\to 0}{\lim }\frac{\log \left(1+\frac{h}{e}\right)}{e\cdot \frac{h}{e}}=\frac{1}{e}\left(\because \underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{x}=1\right]$