# Find the limit as x tends to 0

## Homework Statement

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

## 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 dont want to use L'hospital's rule)

= 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 dont 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.

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??

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...

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

lurflurf
Homework Helper

= 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

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

lurflurf
Homework Helper

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

the second one i agree. but; where did (y^5-1) come from?
the numerator is: y^10-2y^5+1
now that doesnt divide (y-1)^2

lurflurf
Homework Helper

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

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 cant substitute y=1 to find the limit since the denominator would be 0.

lurflurf
Homework Helper

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)]

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

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

i dont like power expansions of high powers.