Why Does Factoring Change the Existence of a Limit in Spivak's Problem?

[Alpharup]

Consider the limit
lim f(x)g(x)
x→a
Spivak has proved that this is equal to lim f(x) multlied by
x→a
lim g(x)
x→a

And also if lim g(x) = k and k≠0,
x→a
Then. lim 1/g(x) = 1/k
x→a

Now the problem arises...
Consider the limit
lim ((x^2)-(a^2))/(x-a)
x→a
It can factorised and written as( taking x-2 from numerator)
lim (x+a)
x→a
Which is nothing but 2a.
Now we can write it the above limit also as
lim(x^2)-(a^2) multiplied by
x→a
lim 1/(x-a)
x→a.

The second limit does not exist because
lim(x-a)=0 and l=0
x→a
So, its reciprocal limit does not exist.
Then can't we say
lim ((x^2)-(a^2))/(x-a) does not exist?
x→a
Where am I wrong in my arguement?

 

[Samy_A]

$\displaystyle\lim_{x\rightarrow a} {f(x)g(x)}=\lim_{x\rightarrow a} {f(x)}.\lim_{x\rightarrow a} {g(x)}$ if the two limits in the RHS exist.

 

[theodoros.mihos]

Take $f(x) = 1/x$ and $g(x) = 1/x^2$. By $x\to\infty$ both limits are zero. Then $\lim_{x\to\infty}\frac{f(x)}{g(x)} = \lim_{x\to\infty}\frac{g(x)}{f(x)}$ even exist or not. Change now reference system to $x\to{x-a}$. Now limits are $-1/a$ and $1/a^2$. Can existence of $\lim_{x\to\infty}\frac{f(x)}{g(x)}$ depends by reference system?

 

[Alpharup]

But, i have not started on 'limit tends to infinity part'. I will start soon.

 

[Samy_A]

But, i have not started on 'limit tends to infinity part'. I will start soon.

Limit to infinity or not is irrelevant. The product rule assumes that the two limits exist.

Let's take a trivial example:
$f(x)=x-a$, $g(x)=\frac{1}{x-a}$
Then $\displaystyle\lim_{x\rightarrow a} {f(x)g(x)}=\displaystyle\lim_{x\rightarrow a} {1}=1$.
But $\displaystyle\lim_{x\rightarrow a} g(x)= \lim_{x\rightarrow a} \frac{1}{x-a}$ doesn't exist, so the expression $\displaystyle\lim_{x\rightarrow a} f(x).\displaystyle\lim_{x\rightarrow a} g(x)$ is not defined.

 

[Alpharup]

$\displaystyle\lim_{x\rightarrow a} {f(x)g(x)}=\lim_{x\rightarrow a} {f(x)}.\lim_{x\rightarrow a} {g(x)}$ if the two limits in the RHS exist.

Yes, i get it. The original condition is the limits should exist. He stated before proving it.
A function f can be written as
f=gb or cd...where g,b, c and d are different functions of x.
let the limit of b as x approaches a, not exist. but for g,c,d the limts exist as x approaces a.
How can we justify the fact that limit of f as f approaches a
is nothing but (limit of c as x approaches a)×( limit of d as x approaches a)
and not (limit of g as x approaches a)×(limit of b as x approaches a)?

 

[Samy_A]

Yes, i get it. The original condition is the limits should exist. He stated before proving it.
A function f can be written as
f=gb or cd...where g,b, c and d are different functions of x.
let the limit of b as x approaches a, not exist. but for g,c,d the limts exist as x approaces a.
How can we justify the fact that limit of f as f approaches a
is nothing but (limit of c as x approaches a)×( limit of d as x approaches a)
and not (limit of g as x approaches a)×(limit of b as x approaches a)?

We justify it by noting that in your last expression, you "multiply" by something that doesn't exist.