Expert: Ahmed Salami - 1/25/2006

Question
lim as x approaches 3 of the square root of (x cubed - 2)=5; PROVE USING EPSILON DELTA METHOD.
I will use E for epsilon and D for delta.
I can get the problem "solved" to the point of:
(E^2-10E+27)^1/3<x<(E^2+10E+27); I don't know what to do to find an equivalent for D (you know, let ? =D)

Thanks

Answer
Hi Donna,
Sorry for the time it took.
To prove that the limit of a function f(x) is equal to L
as x approaches a, we have to show that for all E > 0
there exists a number d(normally d(E)) > 0 such that
|f(x) - L| < E whenever |x - a| < d
Note that E is just a number that measures the closeness
of f(x) to L while d measure that of x to a.
Now, we have to prove that the limit of sqrt(x^3 - 2) as
x approaches 3 is equal to 5. To do this, we know that
L = 5, given E > 0 and d > 0
We have to show that for
|sqrt(x^3 - 2) - 5| < E
x lies in the neighbourhood of 3(i.e very close to 3).
Moving on, we go on simplifying as follows
-E < sqrt(x^3 - 2) - 5 < E
(for we know that when |A| < B, -B < A < B )
Splitting up into two as in
-E < sqrt(x^3 - 2) - 5 and sqrt(x^3 - 2) - 5 < E
5 - E < sqrt(x^3 - 2) and sqrt(x^3 - 2) < 5 + E
(5 - E)^2 < x^3 - 2 and x^3 - 2 < (5 + E)^2
(5 - E)^2 + 2 < x^3 and x^3 < (5 + E)^2 + 2
combining back, we have
(5 - E)^2 + 2 < x^3 < (5 + E)^2 + 2
By simple inspection we can see that foe a very small E,
both (5 - E)^2 and (5 + E)^2 approach 5^2 = 25 and so
27- < x^3 < 27+
3- < x < 3+
(A- means slightly less that A, A+ means slightly more
than A)
And so we have proved that x lies in the neighbourhood
of 3 as required i.e |x - 3| < d(very small)
We could also have given E a small value, say, 0.001
and going back to
(5 - E)^2 + 2 < x^3 < (5 + E)^2 + 2  to get
(4.999)^2 + 2 < x^3 < (5.001)^2 + 2
26.99 < x^3 < 27.01
2.9996 < x < 3.0004
which agrees with what is above.
Note that we are simply showing here that whatever the
value of x we take in this range confirms that
|x - 3| is a very small number, which can be said to be
smaller than d which is also a very small number(say 0.00001).
Pardon me if i've overwritten, i just felt laying out
the solution this much would be helpful.
I hope i have helped. You can always get back to me.
Regards.