Question

I have a question about a derivative proof with logarithm; details in the reply. I have a question about a derivative proof with logarithm; details in the reply. @Mathematics

Answer 1

The following inequality is given:
\[1 - \frac{1}{x} < \ln x < x - 1\]
for every x > 0 and different from 1.
Then the exercise asks to show the following:
\[\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1 \]
but I have no idea on how to start. Can anyone give some hint? I could start from the definition of limit:
\[ \left |\frac{\ln(1+x)}{x} - 1\right | < \epsilon \] whenever \[|x - 0| < \delta\]
but I don't know how to continue.

Answer 2

replace x by 1+x and use squeeze i think gets it in one step

Answer 3

of course i could be wrong, so let me write it out and see if it works

Answer 4

\[\frac{x}{x+1}<\ln(x+1)<x\]

Answer 5

In this case, the limits of both \[\frac{x}{x+1}\] and \[x\] as x approaches zero are 0 respectively.

Answer 6

yup it works. divide by x get
\[\frac{1}{x+1}<\frac{\ln(x+1)}{x}<1\]

Answer 7

This is true, very interesting. Thank you.

Answer 8

yw