How do you find the x values at which #f(x)=x/(x^2-1)# is not continuous, which of the discontinuities are removable?

Answer 1

See below.

This is a rational function.

Rational functions are continuous on their domains.

The domain of this #f# is all reals except #+-1#.
So #f# is discontinuous at #+-1#.
A discontinuity at #a# is removable if #lim_(xrarra)f(x)# exists. But the limit of #f# fails to exist for both #1# and #-1#. So neither discontinuity is removable.
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Answer 2

To find the x values at which f(x) = x/(x^2-1) is not continuous, we need to identify the points where the function is undefined or where the limit does not exist.

The function f(x) is undefined when the denominator, x^2-1, equals zero. Solving x^2-1=0, we find x=1 and x=-1. Therefore, the function is not continuous at x=1 and x=-1.

To determine if these discontinuities are removable, we need to check if the limit of f(x) exists as x approaches these points.

Taking the limit as x approaches 1, we find that the limit is 1/2. Therefore, the discontinuity at x=1 is removable.

Taking the limit as x approaches -1, we find that the limit is also 1/2. Therefore, the discontinuity at x=-1 is removable as well.

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Answer from HIX Tutor

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.

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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