# How do you prove that the function #f(x)=(x-5)/ |x-5|# is continuous everywhere but x=5?

Use the definition of

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To prove that the function f(x) = (x-5)/|x-5| is continuous everywhere except at x=5, we need to show that it satisfies the definition of continuity for all values of x except x=5.

First, let's consider the function for x ≠ 5. In this case, the denominator |x-5| is always positive since x is not equal to 5. Therefore, the function f(x) can be simplified to f(x) = (x-5)/(x-5) = 1, which is a constant function.

A constant function is continuous everywhere, so for all x ≠ 5, f(x) is continuous.

Now, let's examine the behavior of the function at x=5. Since the denominator |x-5| becomes zero at x=5, the function f(x) is undefined at this point. This means that f(x) is not continuous at x=5.

In summary, the function f(x) = (x-5)/|x-5| is continuous everywhere except at x=5, as the function is a constant for all x ≠ 5 and undefined at x=5 due to the denominator becoming zero.

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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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