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

Oct 18, 2016

Limit as x->0 of $\ln \frac{{x}^{2} + 1}{x} = 0$

#### Explanation:

Direct application give $\frac{0}{0}$
So we use l'Hôpital rule $\frac{\ln \left({x}^{2} + 1\right) '}{x '} = \frac{2 x}{{x}^{2} + 1} = \frac{0}{1} = 0$

Oct 18, 2016

l'Hopital's Rule applies.

$\therefore {\lim}_{x \to 0} \ln \frac{2 {x}^{2} + 1}{x} = 0$

#### Explanation:

If we try to evaluate at the limit we obtain:
$\frac{0}{0}$

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:

$\frac{d \left[\ln \left({x}^{2} + 1\right)\right]}{\mathrm{dx}} = \frac{2 x}{{x}^{2} + 1}$

The derivative of the denominator:

$\frac{d \left[x\right]}{\mathrm{dx}} = 1$

Here is our new expression:

${\lim}_{x \to 0} \frac{2 x}{{x}^{2} + 1}$

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

$\therefore {\lim}_{x \to 0} \ln \frac{2 {x}^{2} + 1}{x} = 0$