# Using the definition of convergence, how do you prove that the sequence #lim (n + 2)/ (n^2 - 3) = 0# converges from n=1 to infinity?

To prove that the sequence ( \lim_{n \to \infty} \frac{n + 2}{n^2 - 3} = 0 ) converges as ( n ) approaches infinity, we can use the definition of convergence, which states that for any given positive number ( \epsilon ), there exists a positive integer ( N ) such that for all ( n > N ), the absolute difference between the sequence and its limit is less than ( \epsilon ).

Given the sequence ( \frac{n + 2}{n^2 - 3} ), we need to show that for any ( \epsilon > 0 ), there exists an integer ( N ) such that for all ( n > N ), ( \left| \frac{n + 2}{n^2 - 3} - 0 \right| < \epsilon ).

Since the limit of the sequence is 0, the inequality simplifies to ( \left| \frac{n + 2}{n^2 - 3} \right| < \epsilon ).

Now, observe that for ( n > 1 ), the denominator ( n^2 - 3 ) is always greater than 1, and the numerator ( n + 2 ) increases linearly with ( n ). Therefore, as ( n ) approaches infinity, the sequence approaches zero.

To formalize this, let's choose ( N = \frac{2}{\epsilon} ). Then for all ( n > N ),

[ \left| \frac{n + 2}{n^2 - 3} \right| < \frac{n + 2}{n^2 - 3} < \frac{n + 2}{n^2} < \frac{n + 2}{n} < \frac{n + n}{n} = 2 ]

This implies that ( \left| \frac{n + 2}{n^2 - 3} \right| < 2 ), which is always less than ( \epsilon ).

Therefore, by definition of convergence, the sequence ( \frac{n + 2}{n^2 - 3} ) converges to 0 as ( n ) approaches infinity.

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You apply the deffinition and then simplify

So:

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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