How do you find the excluded value of #3/(x-5)#?

Answer 1

I got #x=5#

You need to consider the value of #x# that makes your fraction impossible to be evaluated; the only option is when #x=5#. This is because you would get #1/0# that you cannot evaluate! So, when you have a fraction with unknown in it (in the denominator) always set the denominator equal to zero, solve it as a normal equation and exclude the value(s) of the unknown you find. For example, in your case: 1) take the denominator: #x-5#; 2) set it equal to zero: #x-5=0#; 3) solve the equation: #x=5#; Now you can exclude the result, #x=5,# from the set of values you can use into your fraction.
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Answer 2

To find the excluded value of 3/(x-5), we need to determine the value(s) of x that would make the denominator, (x-5), equal to zero. Setting the denominator equal to zero and solving for x, we have:

x - 5 = 0

Adding 5 to both sides:

x = 5

Therefore, the excluded value of 3/(x-5) is x = 5.

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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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