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- #1

#### TheAvenger

##### New member

- Feb 24, 2013

- 4

limit x -> 0 log (1 + x)/x = 1

and

limit x -> infinity log (1+x)/x = 0

Thanks in advance!

- Thread starter TheAvenger
- Start date

- Thread starter
- #1

- Feb 24, 2013

- 4

limit x -> 0 log (1 + x)/x = 1

and

limit x -> infinity log (1+x)/x = 0

Thanks in advance!

- Jan 17, 2013

- 1,667

What about using a series expansion , is it allowed ?

- Admin
- #3

- Mar 5, 2012

- 9,016

Hi Avenger!

limit x -> 0 log (1 + x)/x = 1

and

limit x -> infinity log (1+x)/x = 0

Thanks in advance!

Sure.

If you can find a sequence above and another sequence below your sequence, you can "squeeze" you sequence.

For your first limit, we have:

$\qquad x - \dfrac {x^2} 2 \le \log(1+x) \le x \qquad$

If x>0 we get:

$\qquad\dfrac{x - \dfrac {x^2} 2}{x} \le \dfrac{\log(1+x)}{x} \le \dfrac{x}{x} \qquad$

$\qquad1 - \dfrac {x} {2} \le \dfrac{\log(1+x)}{x} \le 1 \qquad$

If x approaches zero from above these expressions will approach 1, so

$\qquad\displaystyle\lim_{x \downarrow 0} \dfrac{\log(1+x)}{x} = 1$

Similarly you can prove that

$\qquad\displaystyle\lim_{x \uparrow 0} \dfrac{\log(1+x)}{x} = 1$

Therefore

$\qquad\displaystyle\lim_{x \to 0} \dfrac{\log(1+x)}{x} = 1$

Can you think of a similar way to do the second limit?

- Thread starter
- #4

- Feb 24, 2013

- 4

Thank you for the help!

- Jan 23, 2013

- 3

- Thread starter
- #6

- Feb 24, 2013

- 4

- Admin
- #7

- Mar 5, 2012

- 9,016

How about $\sqrt x$?

It does not converge to 0, but it is not supposed to.

Neither does $\log(1+x)$.

It is only supposed to increase slower than $x$, but faster than $\log(1+x)$.

- Jan 29, 2012

- 661

Perhaps, the following will be useful for you in the future: $$\lim_{x\to 0}\frac{\log (1+x)}{x}=\lim_{x\to 0}\frac{x+o(x)}{x}=\lim_{x\to 0}\left(1+\frac{o(x)}{x}\right)=1+0=1$$but we haven't covered that yet so I was wondering whether there's some other way to prove these