# a limit problem

• September 13th, 2006, 03:12 PM
coolaak
a limit problem
hi i am a 3rd year bio major and i'm currently taking calc and have to do a HW problem and i cant seem to get the answer... was wondering if you math people could help me with this problem... its only 1 HW problem outta 40 so dont think your helping me do my entire HW.

lim (sqrt(x) - x^2) / 1- sqrtx
x--->1
• September 13th, 2006, 03:26 PM
river_rat
Well, subbing in x=1 into your limit gives the form 0/0 which screams use l'hopital's . So take derivatives of the numerator and denominator and take the limit again.

I get the answer to be 3
• September 13th, 2006, 03:38 PM
coolaak
we haven't learned that rule yet...using mostly algebra to simplify the expression how would u do it
• September 13th, 2006, 03:44 PM
shmoe
Without resorting to l'hopital, write the numerator as:

sqrt(x)*(1-x^(3/2))

The stuff in the parenthesis is a difference of cubes. Namely 1-y^3=(1-y)*(1+y+y^2), except you have an x^(1/2) instead of a y. so your numerator is:

sqrt(x)*(1-sqrt(x))*(1+sqrt(x)+x)

Can you take it from here?
• September 13th, 2006, 03:46 PM
river_rat
Ok, so they want you to do it the hard way...

First times the numerator and denominator by (1 + sqrt(x)) and times out

This gives us the (sqrt(x) + x - x^2 - sqrt(x) x^2)/(1 - x)

We can rearrange this into sqrt(x) (1 - x^2)/(1-x) + x (1 - x)/(1 - x)

which is sqrt(x) (1 + x) + x -> 1 (2) + 1 = 3
• September 14th, 2006, 03:13 AM
mitchellmckain
Re: a limit problem
Of course shmoe's solution is the simplest and most straightforward,....

But, just to be different we could also use the expansion of the sqrt x using the binomial theorem. But this works best if one of the terms in the binomial approaches zero in the limit, so we use the substitution x=1+u, then as x->1, u->0.

So sqrt(x) = sqrt(1+u) = 1 + .5u - .125 u^2 + ...
The advantage is that as u approaches 0 all the higher terms disappear.

instead of substituting back to x it is easier to put everything in u
in which case our limit becomes

lim (1+.5u -.125 u^2 + .... - (u+1)^2)/[-.5 u - .125 u^2 + ...]
u->0

we just need to expand (u+1)^2 = u^2 + 2u + 1

so we get

lim (-1.5 u - 1.125 u^2 + ....)/(-.5 u - .125 u^2 + ...)
u->0

then cancel a factor of u to get

lim (-1.5 - 1.125 u + ....)/(-.5 - .125 u + ...)
u->0

But now numerator and denominator do not go to zero at u=0 so we substitute to get

(-1.5)/(-.5) = 3

You should get used to doing series methods since they have wide application.

But to tell the truth there is a direct relationship between this method above and the LHopital's rule because the binomial expansion here is equivalent to a Taylor's expansion.
• September 15th, 2006, 08:52 AM
coolaak
i just wanted to thank everyone for the help i really do appriciate it... if anyone ever needs help with there bio/chem homework feel free to post it or pm me and i will do my best to help u out.