Epsilon delta surroundings question

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SUMMARY

The discussion centers on proving the inequality \( \frac{1}{|x+1|} > \epsilon \) for every delta (\( d \)). The participants explore the choice of \( x = \min\{2, \frac{1+d}{2}\} \) to satisfy the condition for all \( d \). For \( d > 2 \), \( x \) is defined as \( \frac{1+d}{2} \), while for \( d < 2 \), \( x \) is set to 2. This formulation is derived to ensure the inequality holds universally across the specified delta values.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with epsilon-delta definitions of continuity
  • Basic algebraic manipulation skills
  • Knowledge of inequalities and their properties
NEXT STEPS
  • Study the epsilon-delta definition of limits in calculus
  • Explore examples of proving limits using epsilon-delta arguments
  • Learn about the properties of inequalities in mathematical proofs
  • Investigate the implications of choosing specific values for epsilon and delta in limit proofs
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Students of calculus, mathematics educators, and anyone interested in understanding the formal proofs of limits and continuity in mathematical analysis.

nhrock3
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i need to prove that for every delta
i call d=delta
e=epsilon
1/|x+1|>e
we can choosr any e we want
so they took e=1/4
but because the innqualitty needs to work for every delta
they took x=min{2,(1+d)/2}
for d>2 it takes x= 1+d /2
for d<2 takes x=2
uppon what logic they found this formula x=min{2,(1+d)/2}
 
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nhrock3 said:
i need to prove that for every delta
i call d=delta
e=epsilon
1/|x+1|>e
we can choosr any e we want
so they took e=1/4
but because the innqualitty needs to work for every delta
they took x=min{2,(1+d)/2}
for d>2 it takes x= 1+d /2
for d<2 takes x=2
uppon what logic they found this formula x=min{2,(1+d)/2}

I don't understand the question. You say "for every delta", but then the equation that follows doesn't have a delta in it.
 

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