lim-x-0-log-e-x-1-x-please-help-

Question Number 150759 by nimnim last updated on 15/Aug/21

\underset{x \to 0}{l i m} \frac{l o g (e + x) - 1}{x} = ?

p l e a s e h e l p . .

Answered by Olaf_Thorendsen last updated on 15/Aug/21

\frac{\ln (e + x) - 1}{x} = \frac{\ln e + \ln (1 + \frac{x}{e}) - 1}{x}

= \frac{\ln (1 + \frac{x}{e})}{x} \underset{0}{\sim} \frac{\frac{x}{e}}{x} = \frac{1}{e}

Commented by nimnim last updated on 15/Aug/21

T h a n k y o u s o m u c h ., S i r .

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