Question

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

Answer 1

Limit as x->0 of `ln(x^2+1)/x= 0`

Explanation:

Direct application give `0/0`
So we use l'Hôpital rule `(ln(x^2+1)')/(x')=(2x)/(x^2+1)=0/1=0`

Answer 2

l'Hopital's Rule applies.

`:. lim_(x->0)ln(2x^2 + 1)/x = 0`

Explanation:

If we try to evaluate at the limit we obtain:
`0/0`

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:

`(d[ln(x^2+ 1)])/dx = (2x)/(x^2 + 1)`

The derivative of the denominator:

`(d[x])/dx = 1`

Here is our new expression:

`lim_(x->0) (2x)/(x^2+1)`

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

`:. lim_(x->0)ln(2x^2 + 1)/x = 0`