Question

Use the definition of limits of sequence to give a formal proof of
limₙ→[infinity] (√n²+1-n)-0

Answer 1

To prove the limit of the sequence (√n²+1-n)-0 as n tends to infinity, we can show that for every positive value of ε, there exists a positive integer N such that for every integer n > N, the absolute value of (√n²+1-n)-0 is less than ε. By expanding and simplifying the expression, we can determine that by choosing N = 1/(2ε), we can ensure that (√n²+1-n)-0 approaches 0 as n approaches infinity.

Step-by-step explanation:

To prove the limit of the sequence (√n²+1-n)-0 as n tends to infinity, we need to show that for every positive value of ε, there exists a positive integer N such that for every integer n > N, the absolute value of (√n²+1-n)-0 is less than ε. Let's begin the proof:

1. We have (√n²+1-n)-0 = (√(n²+1)-n)，by expanding the square root.
2. Next, we can simplify (√(n²+1)-n) by multiplying the numerator and denominator by the conjugate, (√(n²+1)+n). This gives us ((n²+1)-n²)/((√(n²+1)+n)).
3. Simplifying further, we have 1/((√(n²+1)+n)).
4. We can observe that 0 < (√(n²+1)+n) < 2n, because (√(n²+1)+n) > (√(n²+1)+1) > (√n²+1+n) > √n²+1+n > 2√n. Here, (√n < n) holds true for all n > 1.
5. Therefore, 0 < 1/((√(n²+1)+n)) < 1/(2n).
6. Now if we choose N = 1/(2ε), for every n > N, we have 0 < 1/((√(n²+1)+n)) < 1/(2n) < ε. This proves that the sequence (√n²+1-n)-0 tends to 0 as n approaches infinity.