Is the expression \(\lim_{x \to 1.5} \sin x = 0.997494986\) with large δ valid?

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SUMMARY

The expression \(\lim_{x \to 1.5} \sin x = 0.997494986\) is invalid when δ is set to 7, as the sine function oscillates between -1 and 1, failing to approach the stated limit directly. The delta-epsilon definition of limits requires that for any ε > 0, there exists a δ > 0 such that if |x - 1.5| < δ, then |sin(x) - L| < ε. The discussion highlights the importance of correctly applying the delta-epsilon proof and clarifies that the limit value of 0.5381 mentioned is incorrect. Understanding these concepts is crucial for accurate limit evaluations in calculus.

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dE_logics
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In a sinusoidal function...suppose the value of δ is very large...then as x approaches any a, the value of f(x) might not approach L directly...or there should not be a direct relation; example -

\lim_{x \to 1.5} sin x = 0.997494986

Where I've stated δ as 7...then if x = 1.5 – 6.9 = -5.4; as x approach 1.5 from -5.4, value of sin x does not directly approach 0.997494986...it fluctuates between 1 to -1 many times before it reaches that value.

My question is...is this expression \lim_{x \to 1.5} sin x = 0.997494986 with δ as 7 valid?
 
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It does not matter "how" x approaches a. The only requirement is that "if |x-a|< delta, then |f(x)- L|< epsilon. It is NOT a matter of x getting "closer and closer to a".

Talking about f(x) changing "as x approaches 1.5", for x distant from 1.5 is completely irrelevant. Given any epsilon> 0, there exist a delta such that if |x- 1.5|< delta, then |sin(x)-0.5381|< epsilon.
 
HallsofIvy said:
It does not matter "how" x approaches a. The only requirement is that "if |x-a|< delta, then |f(x)- L|< epsilon. It is NOT a matter of x getting "closer and closer to a".

Talking about f(x) changing "as x approaches 1.5", for x distant from 1.5 is completely irrelevant. Given any epsilon> 0, there exist a delta such that if |x- 1.5|< delta, then |sin(x)-0.5381|< epsilon.

Oh, ok, I get it...I think.

|sin(x)-0.5381| should not exceed ε if |x- 1.5|< delta.
 
dE_logics said:
Oh, ok, I get it...I think.

|sin(x)-0.5381| should not exceed ε if |x- 1.5|< delta.

Rather the other way around. If |x-1.5| < delta, then |sin(x)-.05381| will be less than epsilon. That's the point of the delta-epsilon proof.
 
We can take either ways.
 
dE_logics said:
We can take either ways.
actually, watch out for the false definition:

for any epsilon > 0, there exists a delta > 0 such that |f(x) - L | < epsilon => |x-a| < delta

this is WRONG. it would be a good exercise disproving this
 
"B if A" is the same as "if A, then B." If you read carefully, you'll notice dE_logics said the right thing (except with an incorrect value for the limit. I don't know where Halls got 0.5381 from...).
 
Moo Of Doom said:
"B if A" is the same as "if A, then B." If you read carefully, you'll notice dE_logics said the right thing (except with an incorrect value for the limit. I don't know where Halls got 0.5381 from...).
Neither do I! I don't know where I got that.
 
Moo Of Doom said:
"B if A" is the same as "if A, then B." If you read carefully, you'll notice dE_logics said the right thing (except with an incorrect value for the limit. I don't know where Halls got 0.5381 from...).

yeah, I noticed that, but it is good practise to disprove the false statement anyway, many functions work under that particular kind of false definition
 

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