Is the square root of 5+1/5 rational or irrational
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From the given table, we have that the lateral limits of f(x) as x -> 3 are different, hence the limit of f(x) does not exist at x = 3.
<h3>What is a limit?</h3>A limit is given by the value of function f(x) as x tends to a value. For the limit to exist, the lateral limits have to be the same, as follows:
In this problem, we have that:
- To the left of x = 3, that is, for values that are less than x = 3, f(x) - > -3.
- To the right of x = 3, that is, for values that are greater than x = 3, f(x) -> 4.
Hence the lateral limits are given as follows:
Since the lateral limits are different, the limit does not exist.
More can be learned about lateral limits at brainly.com/question/26270080
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