Prove the following limit is incorrect using epsilon delta.

[castadiva]

Prove the following limit is incorrect using epsilon delta:

lim x-->4+ (1/4-x) = infinity

The expression within the brackets is f(x), and x is approaching 4 from the right.

I know epsilon delta proofs work this way:

|f(x)-L|<E if |x-a|<d

Now if I try applying this expression, it turns out like this:

|(1/4-x)-inf.|<E; |x-4|<d

...but I'm not sure if I'm doing it right. Suggestions are much appreciated!

 

[castadiva]

Clarification:

f(x) = 1/(4-x). Sorry. I just realized that people could misinterpret this as 1/4-x (as in, one fourth minus x). Hope this helps clear up things!

 

[BigGlenntheHeavy]

\(\displaystyle Is \ that \ \lim_{x\to4^{+}}\frac{1}{4-x} \ or \ \lim_{x\to4^{+}}\frac{1}{4}-x?\)

 

[castadiva]

It's the former. I just made a new post clarifying that. Again, my apologies!

 

[castadiva]

|(1/(4-x))-inf.|<E; |x-4|<d

So I just took a closer look at that... would (1/(4-x))-infinity just be set equal to negative infinity? Or would it be zero? That's the part that confuses me the most.

 

[castadiva]

Does anyone know? Even the slightest hint might help me immensely...

 

[daon]

Limits at infinity have a different definition. Please look that up.

and, well, this should be intuitively obvious at least.1/(4-x) is negative. it is approaching negative infinity.

 

[Aladdin]

\(\displaystyle \lim_{x\to4^{+}}\frac{1}{4-x} \ = \ \lim\frac{1}{0} \ = \ infinity\)

 

[daon]

\(\displaystyle \lim_{x\to4^{+}}\frac{1}{4-x} \ = \ \lim\frac{1}{0} \ = \ infinity\)

lim 1/0 = infinity. Gotcha.

But seriously,

Let \(\displaystyle M < 0\) be arbitrary and let \(\displaystyle \delta=|M^{-1}|\). Then

\(\displaystyle x \in (4,4+\delta) \,\, \implies M^{-1} < 4-x < -M^{-1} \implies \frac{1}{4-x} < M\).

So for any negative number M we pick, we can find a range to the right of 4 where all the x values make it smaller than M.

Castadiva, THIS DOES NOT ANSWER YOUR QUESTION. You need to use the limit definition to show the limit is NOT \(\displaystyle \infinity\).

 

[BigGlenntheHeavy]

\(\displaystyle \lim_{x\to4}\frac{1}{4-x} \ does \ not \ exist.\)

\(\displaystyle The \ limit \ of \ \lim_{x\to4^{+}}\frac{1}{4-x} \ approaches \ negative \ infinity,\)

\(\displaystyle And \ the \ limit \ \lim_{x\to4^{-}}\frac{1}{4-x} \ approaches \ positive \ infinity.\)

\(\displaystyle Note: \ Usually, \ we \ write, \ for \ example, \ \lim_{x\to4^{+}}\frac{1}{4-x} \ = \ -\infty\)

\(\displaystyle This \ is \ for \ the \ sake \ of \ brevity, \ and \ actually \ is \ mathematically \ incorrect,\)

\(\displaystyle as \ a \ number \ can \ never \ equal \ infinity, \ but \ a \ number \ can \ always \ approach \ infinity.\)

\(\displaystyle But, \ this \ is \ all \ redundant, \ as \ you \ want \ a \ rigorous \ proof \ that \ \lim_{x\to4^{+}}\frac{1}{4-x} \ = \ \infty \ is \ incorrect.\)

\(\displaystyle Where \ are \ the \ purists \ when \ one \ is \ needed?\)

 

[galactus]

Try a contradiction proof.

Let me rephrase this:

Assume there is an L as the limit. Then, given an \(\displaystyle {\epsilon}, \;\ \exists \;\ {\delta} > 0\) such that

\(\displaystyle |x-4|<{\delta}\). Then \(\displaystyle \left|L-\frac{1}{x-4}\right|<{\epsilon}\)

Try some small delta values and see.

 

[Aladdin]

\(\displaystyle \lim_{x\to4^{+}}\frac{1}{4-x} \ = \ \infty \ is \ incorrect.\)

My teacher usually write that Lim(number/0) tends to infinity then there's a Vertical Asy. ....

Lim(number/0+) tends to +inf

Lim(number/0-) tends to -inf

Correct ! For sure ...

 

[daon]

\(\displaystyle \lim_{x\to4}\frac{1}{4-x} \ does \ not \ exist.\)

\(\displaystyle The \ limit \ of \ \lim_{x\to4^{+}}\frac{1}{4-x} \ approaches \ negative \ infinity,\)

\(\displaystyle And \ the \ limit \ \lim_{x\to4^{-}}\frac{1}{4-x} \ approaches \ positive \ infinity.\)

\(\displaystyle Note: \ Usually, \ we \ write, \ for \ example, \ \lim_{x\to4^{+}}\frac{1}{4-x} \ = \ -\infty\)

\(\displaystyle This \ is \ for \ the \ sake \ of \ brevity, \ and \ actually \ is \ mathematically \ incorrect,\)

\(\displaystyle as \ a \ number \ can \ never \ equal \ infinity, \ but \ a \ number \ can \ always \ approach \ infinity.\)

\(\displaystyle But, \ this \ is \ all \ redundant, \ as \ you \ want \ a \ rigorous \ proof \ that \ \lim_{x\to4^{+}}\frac{1}{4-x} \ = \ \infty \ is \ incorrect.\)

\(\displaystyle Where \ are \ the \ purists \ when \ one \ is \ needed?\)

I provided a rigorous proof that the limit is negative infinity.

I don't want to DO the proof for the OP.

 

[BigGlenntheHeavy]

What is the OP?

 

[daon]

What is the OP?

Original Post/Poster

 

[BigGlenntheHeavy]

Daon, I don't understand, if you are looking on this board, I assume you want to aid others with your expertise.

Hence, if you know how to solve a problem, why not? Who cares about the OP, whatever that is.

Am I missing something here?

 

[BigGlenntheHeavy]

Daon, I don't understand. If you are looking at this board, then you obviously want to aid those who are less "practiced" than you in regards to your expertise.

What has the OP have to do with anything. Am I missing something here?

 

[daon]

Daon, I don't understand. If you are looking at this board, then you obviously want to aid those who are less "practiced" than you in regards to your expertise.

What has the OP have to do with anything. Am I missing something here?

Chain of events:

The OP asked a question.

I tried to steer him/her in the right direction by mentioning that he/she is using the wrong definition.

I see incorrect "proof" and I add my two cents.

My post is ignored.

And we're still waiting for the poster to clarify.

If they want help, they'll post back. Otherwise I'll just assume they want to "copy and paste."

 

[castadiva]

Hey guys! Thanks so much for your help!

Using some of the suggestions that were on this thread, I came up with the following:

lim x-->4+ 1/(4-x) = infinity, i.e.

for all R > 0 there exists a ?> 0 such that f(x) > R whenever 0 < x - 4 < ?.

So 1/(4 - x) > R

– 1/(x - 4) > R

Multiply both sides by (x - 4)/R which is positive as x approaches 4 from above:

-1/R > (x - 4) or (x - 4) < – 1/R.

We have obtained that 1/(4 - x) > R is satisfied for negative values of (x - 4) but according to definition it should be satisfied for 0 < x - 4 < ?.

Contradiction proves that the limit lim x-->4+ 1/(4-x) = ? is incorrect.

Does that look right to you guys? I personally thought it made sense, but I could be totally wrong haha... math is not my strong point

Again, thanks a ton! Also - this is probably irrelevant but I'm female. Just wanted to clear that up so people wouldn't have to go through the trouble of typing "him/her" every time they made a reference to me. :wink:

 

[daon]

Hey guys! Thanks so much for your help!

Using some of the suggestions that were on this thread, I came up with the following:

lim x-->4+ 1/(4-x) = infinity, i.e.

for all R > 0 there exists a ?> 0 such that f(x) > R whenever 0 < x - 4 < ?.

So 1/(4 - x) > R

You can stop here. Why would this be a contradiction? If you want it to "stick" more, pick any R you want (it has to hold for ALL R, so why not pick something simple like R=1?).

It looks like you know what you're doing for the most part though