How do you find the Limit of #x^2+1# as #x->1# and then use the epsilon delta definition to prove that the limit is L?

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so that:

To find the limit of x^2+1 as x approaches 1, we substitute the value of 1 into the function and evaluate it.

When x=1, we have (1)^2+1 = 2. Therefore, the limit of x^2+1 as x approaches 1 is 2.

To prove this using the epsilon-delta definition, we need to show that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x-1| < delta, then |x^2+1 - 2| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we want to find a delta > 0 such that if 0 < |x-1| < delta, then |x^2+1 - 2| < epsilon.

|x^2+1 - 2| = |x^2-1| = |(x-1)(x+1)| = |x-1||x+1|

Since we are interested in the behavior of x as it approaches 1, we can assume that |x-1| < 1. This implies that -1 < x-1 < 1, which further implies that 0 < x < 2.

Now, let's consider the expression |x^2+1 - 2| = |x-1||x+1|. Since we have assumed that |x-1| < 1, we can also assume that |x-1| < 1/2. This implies that 1/2 < x < 3/2.

Now, let's choose a value for delta. Let delta = min{1/2, epsilon/3}.

If 0 < |x-1| < delta, then we have two cases to consider:

Case 1: If 0 < x-1 < delta, then we have 1/2 < x < 1 + delta. Since delta ≤ 1/2, we have 1/2 < x < 1 + 1/2 = 3/2.

In this case, we can see that |x^2+1 - 2| = |x-1||x+1| < delta * (3/2 + 1) = delta * 5/2 < (epsilon/3) * (5/2) = epsilon * 5/6.

Case 2: If -delta < x-1 < 0, then we have 1 - delta < x < 1/2. Since delta ≤ 1/2, we have 1 - 1/2 < x < 1/2.

In this case, we can see that |x^2+1 - 2| = |x-1||x+1| < delta * (1/2 + 1) = delta * 3/2 < (epsilon/3) * (3/2) = epsilon/2.

Now, let's consider the maximum value between epsilon * 5/6 and epsilon/2. We can see that epsilon * 5/6 > epsilon/2 for any epsilon > 0.

Therefore, we can conclude that if 0 < |x-1| < delta, then |x^2+1 - 2| < epsilon, where delta = min{1/2, epsilon/3}.

Hence, we have proven that the limit of x^2+1 as x approaches 1 is 2 using the epsilon-delta definition.

To find the limit of ( x^2 + 1 ) as ( x ) approaches 1 and then prove it using the epsilon-delta definition, follow these steps:

1. Calculate the limit directly: ( \lim_{x \to 1} (x^2 + 1) ).
2. Evaluate the limit: ( \lim_{x \to 1} (x^2 + 1) = 1^2 + 1 = 2 ).
3. Choose ( \epsilon > 0 ) and find a corresponding ( \delta > 0 ) such that for all ( x ) satisfying ( 0 < |x - 1| < \delta ), ( |(x^2 + 1) - 2| < \epsilon ).
4. Start with ( |x^2 + 1 - 2| < \epsilon ) and simplify to ( |x^2 - 1| < \epsilon ).
5. Factor the expression to ( |x - 1||x + 1| < \epsilon ).
6. We can use the triangle inequality: ( |x - 1||x + 1| \leq |x - 1|(|x| + 1) ).
7. We want to find an upper bound for ( |x| + 1 ). Since ( |x - 1| < \delta ), then ( 1 - \delta < x < 1 + \delta ). Therefore, ( 1 - \delta < |x| < 1 + \delta ), which leads to ( |x| + 1 < 2 + \delta ).
8. Substitute this upper bound into the inequality: ( |x - 1|(|x| + 1) < \delta(2 + \delta) ).
9. We want to choose ( \delta ) so that ( \delta(2 + \delta) < \epsilon ).
10. Solve for ( \delta ) to satisfy the inequality.
11. Finally, choose ( \delta = \min{1, \frac{\epsilon}{3}} ) (for example), then ( |(x^2 + 1) - 2| < \epsilon ) for all ( 0 < |x - 1| < \delta ).