# How do you find the limit of #(e^(-x)) / (1+ln(x))# as x approaches infinity?

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there's not much to do here. breaking them up as they are continuous functions in their domains....

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To find the limit of (e^(-x)) / (1+ln(x)) as x approaches infinity, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get -e^(-x) in the numerator and 1/x in the denominator. As x approaches infinity, both the numerator and denominator tend to zero. Applying L'Hôpital's Rule again, we differentiate the numerator and denominator once more. The derivative of -e^(-x) is e^(-x), and the derivative of 1/x is -1/x^2. As x approaches infinity, the numerator e^(-x) approaches zero, and the denominator -1/x^2 approaches zero as well. Therefore, the limit of (e^(-x)) / (1+ln(x)) as x approaches infinity 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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