What is the limit paradox of the function f(x) = x/{(x-1)2(x-3)}?

[tanzl]

I have a paradox here. Please tell me what is wrong.
I need to prove that $$\lim_{x \rightarrow a-}f(x) = -\infty$$
f(x) = $$\frac{x}{(x-1)^2(x-3)}$$

1st case

For all M , where M is a arbitrary large number, there exists $$\delta$$>0 such that f(x)<M whenever 0 < a-x < $$\delta$$

0 < a-x < $$\delta$$.
a-$$\delta$$ < x < a
Therefore x<a ...(1)

Let |x-a| < $$\delta$$₁ where $$\delta$$₁ is a positive number such that $$\frac{1}{(x-1)^2(x-3)}$$ < P where P is a positive number. ...(2)

f(x) = $$\frac{x}{(x-1)^2(x-3)}$$
< $$\frac{a}{P}$$ ... from (1) and (2)
Let $$\frac{a}{P}$$=M
f(x) < M

Done.

2nd case (The opposite case)

For all M , where M is a arbitrary large number, there exists $$\delta$$>0 such that f(x)>M whenever 0 < a-x < $$\delta$$

As above,
0 < a-x < $$\delta$$.
a-$$\delta$$ < x < a
Therefore x>a-$$\delta$$ ...(1)

Let |x-a| < $$\delta$$₁ where $$\delta$$₁ is a positive number such that $$\frac{1}{(x-1)^2(x-3)}$$ > P where P is a positive number. ...(2)

f(x) = $$\frac{x}{(x-1)^2(x-3)}$$
> $$\frac{a-\delta}{P}$$ ... from (1) and (2)
Let $$\frac{a-\delta}{P}$$=M
f(x) > M

Done.

I derive the both cases with the same arguments but obtain opposite result. What is wrong with my proof?

 

[NoMoreExams]

Haven't read yoru question but you would do \lim_{x \rightarrow a} to give you:

$$\lim_{x \rightarrow a}$$

 

[Dick]

I have a paradox here. Please tell me what is wrong.
I need to prove that $$\lim_{x \rightarrow a-}f(x) = -\infty$$
f(x) = $$\frac{x}{(x-1)^2(x-3)}$$

1st case

For all M , where M is a arbitrary large number, there exists $$\delta$$>0 such that f(x)<M whenever 0 < a-x < $$\delta$$

0 < a-x < $$\delta$$.
a-$$\delta$$ < x < a
Therefore x<a ...(1)

Let |x-a| < $$\delta$$₁ where $$\delta$$₁ is a positive number such that $$\frac{1}{(x-1)^2(x-3)}$$ < P where P is a positive number. ...(2)

f(x) = $$\frac{x}{(x-1)^2(x-3)}$$
< $$\frac{a}{P}$$ ... from (1) and (2)
Let $$\frac{a}{P}$$=M
f(x) < M

Done.

2nd case (The opposite case)

For all M , where M is a arbitrary large number, there exists $$\delta$$>0 such that f(x)>M whenever 0 < a-x < $$\delta$$

As above,
0 < a-x < $$\delta$$.
a-$$\delta$$ < x < a
Therefore x>a-$$\delta$$ ...(1)

Let |x-a| < $$\delta$$₁ where $$\delta$$₁ is a positive number such that $$\frac{1}{(x-1)^2(x-3)}$$ > P where P is a positive number. ...(2)

f(x) = $$\frac{x}{(x-1)^2(x-3)}$$
> $$\frac{a-\delta}{P}$$ ... from (1) and (2)
Let $$\frac{a-\delta}{P}$$=M
f(x) > M

Done.

I derive the both cases with the same arguments but obtain opposite result. What is wrong with my proof?

There's a LOT wrong that 'proof'. It doesn't really prove anything. You can't even BEGIN to prove something about the limit of f(x) as x->a until you know what a is! What is it?

 

[HallsofIvy]

I have a paradox here. Please tell me what is wrong.
I need to prove that $$\lim_{x \rightarrow a-}f(x) = -\infty$$
f(x) = $$\frac{x}{(x-1)^2(x-3)}$$

1st case

For all M , where M is a arbitrary large number, there exists $$\delta$$>0 such that f(x)<M whenever 0 < a-x < $$\delta$$

0 < a-x < $$\delta$$.
a-$$\delta$$ < x < a
Therefore x<a ...(1)

Let |x-a| < $$\delta$$₁ where $$\delta$$₁ is a positive number such that $$\frac{1}{(x-1)^2(x-3)}$$ < P where P is a positive number. ...(2)

f(x) = $$\frac{x}{(x-1)^2(x-3)}$$
< $$\frac{a}{P}$$ ... from (1) and (2)
Let $$\frac{a}{P}$$=M
f(x) < M

Done.

2nd case (The opposite case)

For all M , where M is a arbitrary large number, there exists $$\delta$$>0 such that f(x)>M whenever 0 < a-x < $$\delta$$

As above,
0 < a-x < $$\delta$$.
a-$$\delta$$ < x < a
Therefore x>a-$$\delta$$ ...(1)

Let |x-a| < $$\delta$$₁ where $$\delta$$₁ is a positive number such that $$\frac{1}{(x-1)^2(x-3)}$$ > P where P is a positive number. ...(2)

f(x) = $$\frac{x}{(x-1)^2(x-3)}$$
> $$\frac{a-\delta}{P}$$ ... from (1) and (2)
Let $$\frac{a-\delta}{P}$$=M
f(x) > M

Done.

I derive the both cases with the same arguments but obtain opposite result. What is wrong with my proof?

One problem you have is that you haven't said what a is!
$$\lim_{x \rightarrow a-}f(x) = -\infty$$
is true only if a= 1. where did you use that?

 

[tanzl]

You all are right. It is impossible to define the limit if a is not given.
I confuse this with the case $\lim_{x \rightarrow a} f(x) = a^4$,
where f(x) = x⁴.

https://www.physicsforums.com/showthread.php?t=252170

I think it is correct in x⁴ case because it is continuous and the limit of it when x is approaching "a" is always equals to a⁴ whereas in this question, I can only define that when x is approaching a certain "a", the limit is -$$\infty$$ but generally when a is any numbers, the limit is not $$-\infty$$.

Please help me to check this so that I can make sure that what I am doing is correct.
Prove that $\lim_{x \rightarrow 1}f(x) = 1$ , where $f(x) = \frac{1}{x}$.

For all \epsilon >0 , there exists \delta >0 , such that whenever 0<|x-1|<\delta , then $|f(x)-1|<\epsilon$.

|f(x)-1|
$= |\frac{1}{x} - 1|$
$= |\frac{1-x}{x}|$
$=\frac{|1-x|}{|x|}$
$=|1-x|*\frac{1}{x}$

By letting $\delta = \delta_1$ where $\delta_1$ is an arbitrary positive number such that $\frac{1}{x} < M$ , where M is an arbitrary positive number. (This is the part I am not very sure, my reasoning is there exists $0<|x-1|<\delta_1/$ such that I can get $\frac{1}{|x|} < M$, but how can I make sure that this assumption is always correct)

So, $|f(x)-1| < \delta*M = \epsilon$
where $\delta = min [ \delta_1 ,\frac{M}{\epsilon}]$

 

[tanzl]

Y some of the LATEX is not displayed?

 

[HallsofIvy]

Because you didn't tell it to! You did not include the [ tex ] and [ / tex ] tags. Also do not use the HTML tags ^{and _{in LaTex. use ^ and _ instead.}}

 

[HallsofIvy]

You all are right. It is impossible to define the limit if a is not given.
I confuse this with the case $\lim_{x \rightarrow a} f(x) = a^4$,
where f(x) = x⁴.

https://www.physicsforums.com/showthread.php?t=252170

I think it is correct in x⁴ case because it is continuous and the limit of it when x is approaching "a" is always equals to a⁴ whereas in this question, I can only define that when x is approaching a certain "a", the limit is -$$\infty$$ but generally when a is any numbers, the limit is not $$-\infty$$.

Please help me to check this so that I can make sure that what I am doing is correct.
Prove that $\lim_{x \rightarrow 1}f(x) = 1$ , where $f(x) = \frac{1}{x}$.

For all \epsilon >0 , there exists \delta >0 , such that whenever 0<|x-1|<\delta , then $|f(x)-1|<\epsilon$.

|f(x)-1|
$= |\frac{1}{x} - 1|$
$= |\frac{1-x}{x}|$
$=\frac{|1-x|}{|x|}$
$=|1-x|*\frac{1}{x}$

By letting $\delta = \delta_1$ where $\delta_1$ is an arbitrary positive number such that $\frac{1}{x} < M$ , where M is an arbitrary positive number. (This is the part I am not very sure, my reasoning is there exists $0<|x-1|<\delta_1/$ such that I can get $\frac{1}{|x|} < M$, but how can I make sure that this assumption is always correct)

So, $|f(x)-1| < \delta*M = \epsilon$
where $\delta = min [ \delta_1 ,\frac{M}{\epsilon}]$

You need to be a little more careful. If, say, 0< |x-1|< 1/2, then -1/2< x- 1< 1/2 so
1/2< x< 3/2. Now you can say that 1/x< 1/(1/2)= 2. |f(x)-1|< 2|1- x|.

 

[Almanzo]

Consider the function f(x) = x/{(x-1)²(x-3)}

If x has any value other than 1 or 3, the value of f(x) can be calculated by substituting the value of x in the formula. For example, if x=5, f(x) = f(5) = 5/(4²*2) = 5/32. The limit of f(x), for x approaching 5, whether from above or from below, will also be 5/32, as f is a continuous function on any interval that does not include 1 or 3.

The only nontrivial limits are therefore those for x approaching 1 or 3 from above or from below, and those for x approaching plus or minus infinity (that is, x increasing or decreasing without bound).

First we should realize that f(x) = 0 if and only if x = 0.

For very large |x|, f(x) will be nearly equal to x/x³ = 1/x², which approaches zero from above if x approaches plus or minus infinity. The same should be true for f(x).

For x approaching 1 from below (i.e. x<1 and x<3, but x>0) f(x) will have negative values.
For x approaching 1 from above (i.e. x>1 but x<3) f(x) will also have negative values.
For x approaching 3 from below (i.e. x<3, but x>1) f(x) will still have negative values.
For x approaching 3 from above (i.e. x>3 and x>1) f(x) will have positive values.

If |x-1| < epsilon, |f(x)| will be larger than 1/(3-1)*1/(epsilon)².
If |x-3| < epsilon, |f(x)| will be larger than 1/(3-1)²*1/epsilon.
These absolute values will therefore increase without bounds when epsilon approaches zero.

Therefore f(x) approaches minus infinity if x approaches 1 from above or below, or 3 from below; and f(x) approaches plus infinity if x approaches 3 from above.