Find the limit as x tends to 0

[sara_87]

Homework Statement

find the limit as x tends to 0:
lim x^2/(5th-root(1+5x)-(1+x))

Homework Equations

The Attempt at a Solution

Let y=5th-root(1+5x), so now the limit tends to 1 and the limit is now:

lim (y^5-1)^2/(25(y-1-y^5/5+1/5)

= lim (y^10-2y^5+1)/(25y-25-5y^2+5)

This is where i got stuck.
any help would be very much appreciated.

(Note, i don't want to use L'hospital's rule)

 

[Pere Callahan]

= lim (y^10-2y^5+1)/(25y-25-5y^2+5)

This is where i got stuck.
any help would be very much appreciated.

(Note, i don't want to use L'hospital's rule)

Too bad, it would be really easy with l'hospital's rule, you could plug in y=1+h, converting the limit back to h->0. But expanding the y^10 might be tedious Then you can cancel some h's and see where it goes.

 

[sara_87]

Yes it would be much easier to use L'hospital's rule that's why my teacher prohibitted it.

why would i put y=1+h if i have to expand the power 10?
Surely there must be a simpler way??

 

[Pere Callahan]

You can also do a power series expansion of (1+5x)^1/5 around 0...this would be 1+x+...+o(x³)...The first two terms will cancel, so you will need to keep the third one as well...

 

[sara_87]

Okay, i'll try that but i didnt think i would need to use the taylor expansion.

 

[lurflurf]

= lim (y^10-2y^5+1)/(25y-25-5y^2+5)
should be
= lim (y^10-2y^5+1)/(25y-25-5y^5+5)
simplify to
= lim (y^10-2y^5+1)/(y^5-5y+4)/-5
now factor (hint (y-1)^2 divides the numerator and denominator)
however it may be easier to use
z=(-1+y)^2=[-1+5th-root(1+5x)]^2

 

[sara_87]

are you sure that (y-1)^2 is a factor?

 

[lurflurf]

are you sure that (y-1)^2 is a factor?

yes

(y^5-1)=(y-1)(y^4+y^3+y^2+y+1)
(y^5-5y+4)=(y-1)^2(y^3+2y^2+3y+4)

use synthetic division or long division to see

 

[sara_87]

the second one i agree. but; where did (y^5-1) come from?
the numerator is: y^10-2y^5+1
now that doesn't divide (y-1)^2

 

[lurflurf]

y^10-2*y^5+1=(y^5-1)^2

 

[sara_87]

oh right, sorry.
but still, (y^5-1)^2 divides y-1 and the denominator divides (y-1)^2
when we cancel this out we get:
lim (y^4+y^3+y^2+y+1)/(-5)(y-1)(y^3+2y^2+3y+4)

so we still can't substitute y=1 to find the limit since the denominator would be 0.

 

[lurflurf]

since the numerator is square all the factors appear twice
y^10-2*y^5+1=(y^5-1)^2=(y-1)^2(y^3+2y^2+3y+4)^2

lim (y^10-2y^5+1)/(y^5-5y+4)/-5
lim (y^5-1)^2/(y^5-5*+4)/-5
lim [(y^4+y^3+y^2+y+1)(y-1)]^2/[(-5)(y-1)^2(y^3+2y^2+3y+4)]
so that your
lim (y^4+y^3+y^2+y+1)/(-5)(y-1)(y^3+2y^2+3y+4)
should be
lim (y^4+y^3+y^2+y+1)^2/[(-5)(y^3+2y^2+3y+4)]

 

[Pere Callahan]

Is this really easier than using a power expansion of this fifth root...?

 

[sara_87]

Oh right, that makes sense wat a silly mistake i made!
Thanks.

i don't like power expansions of high powers.