# How do you find the limit of # [1/(3+X)]- (1/3) ÷ X# as x->0?

Assuming you meant:

Multiply the fraction by the inner denominators:

Now we can evaluate the limit:

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To find the limit of the expression [1/(3+X)]- (1/3) ÷ X as x approaches 0, we can simplify the expression first.

First, let's simplify the expression [1/(3+X)]- (1/3) ÷ X:

[1/(3+X)]- (1/3) ÷ X = [1/(3+X)] - (1/3) * (1/X)

Next, let's find a common denominator for the two fractions:

[1/(3+X)] - (1/3) * (1/X) = [1/(3+X)] - (1/3) * (X/(X))

Simplifying further:

[1/(3+X)] - (1/3) * (X/(X)) = [1/(3+X)] - (X/3X)

Now, let's combine the fractions:

[1/(3+X)] - (X/3X) = [1/(3+X)] - (X/3X) * (3/3)

Simplifying again:

[1/(3+X)] - (X/3X) * (3/3) = [1/(3+X)] - (3X/9X)

Combining the fractions:

[1/(3+X)] - (3X/9X) = [1/(3+X)] - (1/3)

Now, we can find the limit as x approaches 0:

lim(x->0) [1/(3+X)] - (1/3) = 1/3 - 1/3 = 0

Therefore, the limit of the expression [1/(3+X)]- (1/3) ÷ X as x approaches 0 is 0.

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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 #[1/ln(x)] - [1/x-1]# as x approaches 1?

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