f (x) ={ 1x−1 if x 6 = 1;0 if x = 1.(a) Does limx→5f (x) exist? If so, what is it? Try and establish thevalidity of your answer formally using an epsilon-delta argument.If it exists, does it equal f (5)? Is this function continuous at thepoint x = 5?(b) Does limx→1f (x) exist? If so, what is it? Try and establish thevalidity of your answer formally using an epsilon-delta argument.If it exists, does it equal f (1)? Is this function continuous at thepoint x = 1?(c) Is this function continuous?2. Consider the functionf (x) ={x2 if x 6 = 0;−50 if x = 0.2
Question
f (x) ={ 1x−1 if x 6 = 1;0 if x = 1.(a) Does limx→5f (x) exist? If so, what is it? Try and establish thevalidity of your answer formally using an epsilon-delta argument.If it exists, does it equal f (5)? Is this function continuous at thepoint x = 5?(b) Does limx→1f (x) exist? If so, what is it? Try and establish thevalidity of your answer formally using an epsilon-delta argument.If it exists, does it equal f (1)? Is this function continuous at thepoint x = 1?(c) Is this function continuous?2. Consider the functionf (x) ={x2 if x 6 = 0;−50 if x = 0.2
Solution
(a) The limit as x approaches 5 of f(x) does exist. Since x is not equal to 1, we use the function 1x - 1. Substituting 5 into the function gives us 1*5 - 1 = 4. Therefore, limx→5f(x) = 4.
To establish this formally using an epsilon-delta argument, we need to show that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 5| < δ, then |f(x) - 4| < ε. Since f(x) = 1x - 1, |f(x) - 4| = |1x - 5|. So if we choose δ = ε, then whenever 0 < |x - 5| < δ, we have |f(x) - 4| = |1x - 5| < ε.
Since f(5) = 4, the function is continuous at x = 5.
(b) The limit as x approaches 1 of f(x) does not exist. This is because the function is defined differently for x = 1 (f(x) = 0) and x ≠ 1 (f(x) = 1x - 1). Therefore, the left-hand limit and the right-hand limit as x approaches 1 are not equal, so the limit does not exist.
(c) Since the limit as x approaches 1 does not exist, the function is not continuous at x = 1. Therefore, the function is not continuous.
- For the function f(x) = {x^2 if x ≠ 0; -50 if x = 0}, the limit as x approaches 0 does not exist because the function is defined differently for x = 0 and x ≠ 0. Therefore, the function is not continuous at x = 0.
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