Evaluating Limits/Limit of a Function

KIVES

I've
a couple of problems that have been bewitching me this past 24 hours.
The first one looks relatively handy and probably is in truth. The
second one is the rock on which I've really floundered. Anyway, I'll
trot them out...
EVALUATE Lim x>a (x-a)^3/x^2 - a^2
= (x - a)(x^2 + ax + a^2)/(x-a)(x+a)
=Having cancelled out both (x - a)'s, I'm left with..
=x^2 + ax + a^2/x + a
I substitute a into the equation as lim x>a
=a^2 + a^2 + a^2/a + a
= 3a^2/2a
Where to go from here or I may have already gone wrong. Incidentally,
the
answer is supposed to be 0

QUESTION 2: -
Write down lim n>infinity (r)^n, if [r]<1 (These should be
straight lines either side of the r, as in modulus?)
Hence find lim n tending toward infinity a(1 - r^n)/1 - r, if [r]<1
Fluffed around with this one. Got the first part but the second bit left
me furrowing my brow...again, it may be a synch, just my powers of
interpretation failed me...
Any suggestions as to the 2 questions above would be great...

Fremen

KIVES wrote: »

I've
a couple of problems that have been bewitching me this past 24 hours.
The first one looks relatively handy and probably is in truth. The
second one is the rock on which I've really floundered. Anyway, I'll
trot them out...
EVALUATE Lim x>a (x-a)^3/x^2 - a^2
= (x - a)(x^2 + ax + a^2)/(x-a)(x+a)
=Having cancelled out both (x - a)'s, I'm left with..
=x^2 + ax + a^2/x + a
I substitute a into the equation as lim x>a
=a^2 + a^2 + a^2/a + a
= 3a^2/2a
Where to go from here or I may have already gone wrong. Incidentally,
the
answer is supposed to be 0

You factorised x^3 - a^3, but the question says (x - a)^3. Easy mistake to make. It's so common it has its own name: the freshman's dream.

QUESTION 2: -
Write down lim n>infinity (r)^n, if [r]<1 (These should be
straight lines either side of the r, as in modulus?)
Hence find lim n tending toward infinity a(1 - r^n)/1 - r, if [r]<1
Fluffed around with this one. Got the first part but the second bit left
me furrowing my brow...again, it may be a synch, just my powers of
interpretation failed me...
Any suggestions as to the 2 questions above would be great...

So what'd you get for the limit of r? No need for absolute value signs. To see this, write out the first few terms of (-1/2)^n.

If you know the limit of r^n, all you need to do is sub it into the second expression.

KIVES

Having looked at the answer at the back of the book, I kind of worked toward that figure. The answer to the first part is 0 so having squared both parts of [r]<1, I got r=square root of 1...which could be positive or minus 1. Again, I'm probably wrong on that bit but I assumed values between that assumption to be true. Imagining such values to be tending toward zero, I subbed various powers in for (1/2)^1,2,3 etc...which naturally served my assumption well. Yet you have -1/2 as your chosen value for r?
Again, I must admit I was flailing around at this point and despite being aware that it was probably an easy enough question, the auld brain was fried...

KIVES

Right, so having fallen foul of the 'Freshman's Dream' trapdoor...I re-multiplied the top and got x^3 - ax^2 + 3a^2x - a^2/x^2 - a^2...Is this the point at which I substitute 'a' into the expression to get: -
= a^3 - a^3 + 3a^3 - a^3/a^2 - a^2
=3a^3 - a^3/ 0
=2a^3/0....Answer is zero?
I know the book gives it as zero but am I remotely close to having approached it in the right way?
Or if I had simply left it as (x - a)(x^2 - 2ax + a^2)/(x - a)(x + a)
Then I would have ended up with x^2 - 2ax + a^2/(x + a), which,when having factorised the top would leave me with: -
(x - a)(x - a)/(x + a)
Substitute 'a' into expression and I'm left with 0/2a...which is Zero
One of these approaches has to be right surely?

Fremen

I think I misunderstood your first post, sorry. I was trying to show you that any number that satisfies

r<1
and r>-1

at the same time is fine, so I picked r = -1/2. I could have picked r = 1/7.782197, or r = -.999, or whatever, as long as both of the above hold at the same time.

You said the answer to the first part is 0. Is that according to the back of the book, or according to you?

If you multiply two numbers that are smaller than one together, the result is smaller in absolute value than either of the original numbers.

For instance, think of (1/7)(-1/9) = -1/49

Taking the absolute value of all three, we see that 1/49 is much smaller than either 1/7 or 1/9. Hence, it is much closer to 0 (HINT!).

KIVES wrote: »

Right, so having fallen foul of the 'Freshman's Dream' trapdoor...I re-multiplied the top and got x^3 - ax^2 + 3a^2x - a^2/x^2 - a^2...

There's your mistake. -ax^2 should be -3ax^2, which would have given you 0/0. The other method is fine.

As a side note, if you've got a load of terms organised into brackets, it's rarely a good idea to multiply them out. The solution to a problem will almost never depend on you doing this.

LeixlipRed

Also, it's clear the OP isn't aware that division by zero is undefined.