Verification of the Limit:
To verify the limit limπ₯β1(π₯β1)2=0, we can use the definition of a limit. The definition of a limit states that for a function π(π₯), as x approaches a value π, if for every π>0 there exists a πΏ>0 such that whenever 0<|π₯βπ|<πΏ, then |π(π₯)βπΏ|<π, where L is the limit.
Step-by-Step Verification:
Given the limit limπ₯β1(π₯β1)2=0, we need to show that for any π>0, there exists a πΏ>0 such that whenever 0<|π₯β1|<πΏ, then |(π₯β1)2β0|<π.
Letβs proceed with the verification:
- Start with:|(π₯β1)2β0|=|(π₯β1)(π₯β1)|=|π₯β1||π₯β1|=|π₯β1|2
- We want to show that this expression can be made less than any positive value of epsilon, i.e., we want to show:|(π₯β1)2|<π
- This implies:|π₯β1|2<π
- To simplify further, consider:|π₯β1|2=(π₯β1)2
- Now, letβs choose πΏ=πππ(1,π). Then if:0<|π₯β1|<πππ(1,π)π₯β1<πππ(1,π)(π₯β1)2<(πΏ)2
- Since we chose πΏ=πππ(1,π) and we have shown that:(π₯β1)2<(πΏ)2This implies:|(π₯β1)2β0|=|(π₯β1)2|<(πΏ)2β€π
Therefore, by choosing appropriate values for delta and epsilon and showing that the condition holds true, we have verified that the limit is indeed equal to zero.
Conclusion:
The verification using the definition of limits confirms that: ππππ₯β1(π₯β1)2=0.
Verification of the Limit:
To verify the limit limπ₯β1(π₯β1)2=0, we can use the definition of a limit. The definition of a limit states that for a function π(π₯), as x approaches a value π, if for every π>0 there exists a πΏ>0 such that whenever 0<|π₯βπ|<πΏ, then |π(π₯)βπΏ|<π, where L is the limit.
Step-by-Step Verification:
Given the limit limπ₯β1(π₯β1)2=0, we need to show that for any π>0, there exists a πΏ>0 such that whenever 0<|π₯β1|<πΏ, then |(π₯β1)2β0|<π.
Letβs proceed with the verification:
- Start with:|(π₯β1)2β0|=|(π₯β1)(π₯β1)|=|π₯β1||π₯β1|=|π₯β1|2
- We want to show that this expression can be made less than any positive value of epsilon, i.e., we want to show:|(π₯β1)2|<π
- This implies:|π₯β1|2<π
- To simplify further, consider:|π₯β1|2=(π₯β1)2
- Now, letβs choose πΏ=πππ(1,π). Then if:0<|π₯β1|<πππ(1,π)π₯β1<πππ(1,π)(π₯β1)2<(πΏ)2
- Since we chose πΏ=πππ(1,π) and we have shown that:(π₯β1)2<(πΏ)2This implies:|(π₯β1)2β0|=|(π₯β1)2|<(πΏ)2β€π
Therefore, by choosing appropriate values for delta and epsilon and showing that the condition holds true, we have verified that the limit is indeed equal to zero.
Conclusion:
The verification using the definition of limits confirms that: ππππ₯β1(π₯β1)2=0.