<1> Lim ln(2x-1) / sin (pi)x

x->1

<2>Lim [((x+1)/ln(x+1)) -1/x)]

x->0

Printable View

- March 1st 2006, 06:07 AMbobby77plase help
<1> Lim ln(2x-1) / sin (pi)x

x->1

<2>Lim [((x+1)/ln(x+1)) -1/x)]

x->0 - March 1st 2006, 06:57 AMCaptainBlackQuote:

Originally Posted by**bobby77**

As:

,

and

we need to use L'Hopital's rule to evaluate the limit so we need:

,

and:

.

Then L'Hopital's rule tells us that:

.

Finally checking with numerical experiments indicates that this is right.

RonL - March 1st 2006, 07:31 AMCaptainBlackQuote:

Originally Posted by**bobby77**

This is indeterminate as it is so rearrange:

.

Now apply L'Hopital's rule:

which is indeterminate, so we need the second derivatives:

Which again checks out numericaly

RonL