# How do you find the limit #ln(x^2+1)/x# as #x->0#?

Limit as x->0 of

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l'Hopital's Rule applies.

This means that l'Hopital's Rule applies.

To apply l'Hopital's Rule, you, compute the derivative of numerator, compute the derivative of the denominator, and then reassemble the two derivatives into a new fraction.

The derivative of the numerator:

The derivative of the denominator:

Here is our new expression:

l'Hopital's Rule states that the limit of our new expression goes to the limit as the original expression

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To find the limit of ln(x^2+1)/x as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (2x)/(x^2+1) for the numerator and 1 for the denominator. Evaluating the limit of this expression as x approaches 0, we find that it equals 0. Therefore, the limit of ln(x^2+1)/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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- What is the limit as x approaches 0 of #x/arctan(4x)#?
- How do you find vertical asymptotes using limits?
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