 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Getting Powerful With Euler (Posted on 2009-07-28) Evaluate:

(1 + y)1/y - e
Limit   ----------------
y → 0          y

Note: e denotes the Euler’s number.

 See The Solution Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) exact solution | Comment 2 of 3 | from the definition of eulers number we have
lim  (1+y)^(1/y)=e
y->0

so from this both the numerator and denomonator
tend to zero in the limit so we can use
l'hospital's rule to get
(1+y)^(1/y)-e
lim  -------------=
y->0       y

lim  [d/dy (1+y)^(1/y)-e]=
y->0

lim  d/dy e^[ln(y+1)/y]=
y->0

lim (y+1)^(1/y)*(y-(y+1)ln(y+1))/[y^2(y+1)]
y->0

now we know that
lim (y+1)^(1/y)=e
y->0

so we have

lim  (y+1)^(1/y)*(y-(y+1)ln(y+1))/[y^2(y+1)]=
y->0

e*lim (y-(y+1)ln(y+1))/(y^2(y+1))
y->0

and again we can use l'hospital's rule to get

e*lim -ln(y+1)/(y(3y+2))
y->0

and using l'hospital on last time we get

e*lim -1/[(y+1)*(6y+2)]
y->0

which can easily be seen to be -e/2
which agrees with numerical approximations that I got
on my calculator.

 Posted by Daniel on 2009-07-28 13:02:43 Please log in:
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