# plase help

• Mar 1st 2006, 06:07 AM
bobby77
plase help
<1> Lim ln(2x-1) / sin (pi)x
x->1

<2>Lim [((x+1)/ln(x+1)) -1/x)]
x->0
• Mar 1st 2006, 06:57 AM
CaptainBlack
Quote:

Originally Posted by bobby77
<1> Lim ln(2x-1) / sin (pi)x
x->1

What is:

$
\lim_{x \rightarrow 1} \frac{\ln(2x-1)}{\sin(\pi x)}?
$

As:

$
\lim_{x \rightarrow 1} \ln(2x-1)=\ln(1)=0
$
,

and

$
\lim_{x \rightarrow 1} \sin(\pi x)=0
$

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

$
\lim_{x \rightarrow 1} \frac{d}{dx}\ln(2x-1)=\lim_{x \rightarrow 1} \frac{2}{2x-1}=2
$
,

and:

$
\lim_{x \rightarrow 1} \frac{d}{dx}\sin(\pi x)=\lim_{x \rightarrow 1}\pi \cos(\pi x)=- \pi
$
.

Then L'Hopital's rule tells us that:

$
\lim_{x \rightarrow 1} \frac{\ln(2x-1)}{\sin(\pi x)}=\frac{\lim_{x \rightarrow 1} \frac{d}{dx}\ln(2x-1)}{\lim_{x \rightarrow 1} \frac{d}{dx}\sin(\pi x)}=-\frac{2}{\pi}
$
.

Finally checking with numerical experiments indicates that this is right.

RonL
• Mar 1st 2006, 07:31 AM
CaptainBlack
Quote:

Originally Posted by bobby77
<2>Lim [((x+1)/ln(x+1)) -1/x)]
x->0

What is:

$
\lim_{x \rightarrow 0} \left\{ \frac{x+1}{\ln(x+1)}-\frac{1}{x} \right\}?
$

This is indeterminate as it is so rearrange:

$
\lim_{x \rightarrow 0} \left\{ \frac{x(x+1)-\ln(x+1)}{x \ln(x+1)} \right\}
$
.

Now apply L'Hopital's rule:

$
\lim_{x \rightarrow 0} \left\{ \frac{x(x+1)-\ln(x+1)}{x \ln(x+1)} \right\}=$
$
\frac{\lim_{x \rightarrow 0} \frac{d}{dx}\{x(x+1)-\ln(x+1)\}}{\lim_{x \rightarrow 0} \frac{d}{dx} \{x \ln(x+1)\} }=\frac{0}{0}
$

which is indeterminate, so we need the second derivatives:

$
\lim_{x \rightarrow 0} \left\{ \frac{x(x+1)-\ln(x+1)}{x \ln(x+1)} \right\}=$
$
\frac{\lim_{x \rightarrow 0} \frac{d^2}{dx^2}\{x(x+1)-\ln(x+1)\}}{\lim_{x \rightarrow 0} \frac{d^2}{dx^2} \{x \ln(x+1)\} }=\frac{3}{1}=1.5
$

Which again checks out numericaly

RonL