wachiraj862Lv1

27 Jun

Verification of the Limit:

To verify the limit lim𝑥→1(𝑥−1)2=0, we can use the definition of a limit. The definition of a limit states that for a function 𝑓(𝑥), as x approaches a value 𝑐, if for every 𝜖>0 there exists a 𝛿>0 such that whenever 0<|𝑥−𝑐|<𝛿, then |𝑓(𝑥)−𝐿|<𝜖, where L is the limit.

Step-by-Step Verification:

Given the limit lim𝑥→1(𝑥−1)2=0, we need to show that for any 𝜖>0, there exists a 𝛿>0 such that whenever 0<|𝑥−1|<𝛿, then |(𝑥−1)2−0|<𝜖.

Let’s proceed with the verification:

1. Start with:|(𝑥−1)2−0|=|(𝑥−1)(𝑥−1)|=|𝑥−1||𝑥−1|=|𝑥−1|2
2. We want to show that this expression can be made less than any positive value of epsilon, i.e., we want to show:|(𝑥−1)2|<𝜖
3. This implies:|𝑥−1|2<𝜖
4. To simplify further, consider:|𝑥−1|2=(𝑥−1)2
5. Now, let’s choose 𝛿=𝑚𝑖𝑛(1,𝜖). Then if:0<|𝑥−1|<𝑚𝑖𝑛(1,𝜖)𝑥−1<𝑚𝑖𝑛(1,𝜖)(𝑥−1)2<(𝛿)2
6. Since we chose 𝛿=𝑚𝑖𝑛(1,𝜖) and we have shown that:(𝑥−1)2<(𝛿)2This implies:|(𝑥−1)2−0|=|(𝑥−1)2|<(𝛿)2≤𝜖

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: 𝑙𝑖𝑚𝑥→1(𝑥−1)2=0.