Math Problem Statement
Use the ε-δ definition of the limit for functions to show lim (x → 1) (x^2 - 3x) = -2.
Solution
To prove limx→1(x2−3x)=−2 using the ε-δ definition of the limit, we need to follow these steps:
ε-δ Definition
The statement limx→1f(x)=−2 means:
For every ε>0, there exists a δ>0 such that whenever 0<∣x−1∣<δ, it follows that ∣f(x)−(−2)∣<ε.
Here f(x)=x2−3x.
Proof Steps
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Start with the inequality ∣f(x)−(−2)∣<ε:
∣f(x)−(−2)∣=∣(x2−3x)+2∣=∣x2−3x+2∣.
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Factorize the expression x2−3x+2:
x2−3x+2=(x−1)(x−2).
Therefore,
∣x2−3x+2∣=∣(x−1)(x−2)∣.
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Simplify the inequality:
∣(x−1)(x−2)∣<ε.
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Control the term ∣x−2∣:
When x is close to 1, say ∣x−1∣<1, it follows that:
∣x−2∣=∣(x−1)−1∣≤∣x−1∣+∣1∣<1+1=2.
So, ∣x−2∣<2 if ∣x−1∣<1.
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Bound ∣(x−1)(x−2)∣:
Using ∣x−2∣<2, we can write:
∣(x−1)(x−2)∣=∣x−1∣⋅∣x−2∣<∣x−1∣⋅2.
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Choose δ:
To ensure ∣(x−1)(x−2)∣<ε, we require:
∣x−1∣⋅2<ε⟹∣x−1∣<2ε.
Therefore, choose δ=min(1,2ε).
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Verify:
If 0<∣x−1∣<δ, then:
∣(x−1)(x−2)∣<∣x−1∣⋅2<2ε⋅2=ε.
Thus, ∣f(x)−(−2)∣<ε, completing the proof.
Conclusion
We have shown that for any ε>0, choosing δ=min(1,2ε) satisfies the condition of the ε-δ definition of the limit. Hence, limx→1(x2−3x)=−2.
Do you want further clarification on any step or more examples? 😊
Related Questions
- How can you interpret the ε-δ definition geometrically?
- What changes if the limit were x→c where c=1?
- How would you prove a one-sided limit using the ε-δ definition?
- What happens if f(x) is not continuous at the point of the limit?
- Can you use the ε-δ definition to prove the limit of a rational function?
Tip
Always start by simplifying the function and bounding terms before choosing δ. This helps streamline your proof!