# How do you evaluate #lim_(xtooo) (5 - x^(1/2))/(5 + x^(1/2))# ?

Use L'Hôpital's rule

You apply the rule by differentiating the number and the denominator:

According to the rule, so goes the original limit:

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I got:

If you try directly you get:

We can use de L'Hospital Rule (deriving top and bottom to get):

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To evaluate the limit of (5 - x^(1/2))/(5 + x^(1/2)) as x approaches infinity, we can use the concept of limits at infinity.

First, we simplify the expression by multiplying both the numerator and denominator by the conjugate of the denominator, which is (5 - x^(1/2)). This gives us:

[(5 - x^(1/2))(5 - x^(1/2))]/[(5 + x^(1/2))(5 - x^(1/2))]

Simplifying further, we get:

(25 - 2x^(1/2) + x)/(25 - x)

Now, as x approaches infinity, the term 2x^(1/2) becomes insignificant compared to x. Therefore, we can ignore it in the numerator. Similarly, the term -x becomes insignificant compared to 25 in the denominator.

Thus, the simplified expression becomes:

25/25

Which simplifies to:

1

Therefore, the limit of (5 - x^(1/2))/(5 + x^(1/2)) as x approaches infinity is 1.

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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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- How do you find the limit of #e^(x-x^2)# as #x->oo#?

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