Posted by Jonathan Burros on September 11, 2002 at 15:48:24:
In Reply to: Checking my work posted by Mr. A on September 10, 2002 at 11:19:16:
: lim (1+(4/x))^x
: x->0
: ln y = X ln (1 + (4/x))
: [ln (1 + (4/x))]/(1/x)
: L'Hopitals
: { 1/[1+(4/x)] * -4x^-2 } / (-x^-2)
: rearrange
: 4x/(x+4) --> 0 at x-->0
: ln y = 0
: e^0 ==== 1
: 9-11-2002
: The line
: 4x/(x+4) --> 0 at x-->0
: should say
: 4x/(x+4) --> 0 as x-->0
: Define y = (1 + 4/x)^x
: ln y = x ln(1 + 4/x)
: = [ln(1 + 4/x)]/(1/x)
:By LHospital's Rule
: limit as x-->0 ln y
:= limit as x-->0 [ln(1 + 4/x)]/(1/x)
:=limit as x-->0{ 1/[1+(4/x)] * -4x^-2 } / (-x^-2)
:= limit as x-->0 4/[1+(4/x)]
:= limit as x-->0 (4x)/(x+4)
:=0
: You don't say ln y=0
: You mean limit of ln y as x-->0 equals 0
: So the limit of y as x-->0 equals e^0
:which equals 1
: Other than that your work is correct