Calculus Limit Problem?

Question

Let me tell you the problem first:

It's the limit as x approaches a, and the top of the expression is square root of (2*a^3*x - x^4) minus a*cubed root(a^2*x), and all of that is over the bottom, which is: a minus the fourth root of (a*x^3)

To help anyone out, it has this whole history before it with something about l'Hospital's Rule, something about how it was one of the first ones if one of his first books or something. So I did Hospital's Rule, and then plugged in a for x, and ended up getting 16a/9.

As a side note, when I type " * ", i mean "times"

Can anyone help?

By the way, this is a college problem from my calculus textbook

what do u mean the limit goes to 0? the problem is the limit as x approaches a...

Answer 1

I believe that the other people who answered this question before me may have misunderstood your question. Substituting a for x into this problem does yield the indeterminant form of 0/0. We get:

sqrt(2*a^3*a - a^4) - a*cubed root (a^2*a) / a - fourth root (a*a^3)

Which simplifies to;

sqrt (2*a^4 - a^4) - a*cubed root (a^3) / a - fourth root (a^4)

= sqrt (a^4) - a^2 / a - a

which is obviously = to 0/0.

Therefore we can apply L'Hopital's rule to this problem and it does infact give the answer 16a/9. since we get:

lim as x approches a of 1/2[(2a^3*x -x^4)^(-1/2)] * (2a^3 - 4x^3) - (1/3)[(a^3)(a^2*x)^(-2/3)] / (-1/4)[(ax^3)^(-3/4)] * (3ax^2)

Now we substitute a for x and get:

(1/2)[(2a^4 - a^4)^(-1/2)] * (2a^3 - 4a^3) - (1/3)(a^3)[(a^3)^(-2/3)] /
(-1/4)[(a^4)^(-3/4)] * (3a^3)

which simplifies too,

[(a^4)^(-1/2)] * (-a^3) - (1/3)(a^3)[(a^3)^(-2/3)] / (-3/4)(a^3)[(a^4) ^(-3/4)]

which is equal too:

-a - (1/3)a / (-3/4) = (4/3)(4a/3) = 16a/9

Actually, this problem was used by the Marquis de L'Hopital to demonstrate his rule (which was actually bought from the swiss mathematician Johann Bernoulli), in his book Analyse des Infiniment Petits (Analysis of The Infinitly Small, for all you English speaking people). This book is credited as the first ever Calculus textbook published (1696).

Answer 2

ion599 is correct:

As x approaches a, your limit becomes:

a^2-a^2 / -a = 0, if I understand your initial equation correctly. Therefore:

lim x->a [sqrt(2a^3 * x - x^4) - a*(cubicroot(a^2 * x))]/(-fourth root)(a*x^3) =

[sqrt (2a^3*a-a^4)-a*(cubicroot(a^2*a))]/(-fourth root(a^4)) =

a^2-a^2 / -a = 0.

The Hopital Rule would apply if your initial limit became 0/0, which is indetermined. If such was the case, then the limit can be obtained by:

Calculating G=(derivative of numerator) / (derivative of denominator) and calculating lim x->a of G.

Answer 3

Actually, what help are you looking for?
You don't agree that the answer is 16a/9?
Or you want to find out why L'Hospital's rule works?

Below is my rough thoughts, and they are not the formal explanation.

When two functions, f and g, are zero when x=a, then what is the ratio of them, f(x)/g(x) where x is slightly bigger than a? Since f'(a) = gradient at a ~= (diff of y values)/(diff of x values) = (f(x)-f(a))/(x-a), we may estimate f(x) by
(x-a)f'(a) + f(a)
= (x-a)f'(a) + 0

Thus, f(x)/g(x) can be estimated by
= ((x-a)f'(a)) / ((x-a)g'(a))
= f'(a) / g'(a)

Answer 4

note that 4x - x^2 = x(4 - x) = x(2 + sqrt(x))(2 - sqrt(x)) now sub in to give: lim (2 - sqrt(x)) / [x(2 + sqrt(x))(2 - sqrt(x)) (2 - sqrt(x)) / (2 - sqrt(x)) cancels to 1 as long as x =/= 4, and is a removable discontinuity you can substitute in 4 for x in what's left to find the limit 1 / [x(2 + sqrt(x)] as x ==> 4 is 1 / [4(2 + 2)] = 1/16 limit as x ==> 4 = 1/16 when you initially substitue 4 for x, you get 0 / 0, which is indeterminate, telling you that you can look for factors of 0 to cancel..

Answer 5

the limit goes to zero

Answer 6

Holy crap I got a headache just reading all that math stuff. I hate math...its evil. But, good luck to you!

Answer 7

if a-1b=xb+af=

cdx1