What is the limit as x approaches 0 of (ln(x^2+1))/x ?

The limit as x approaches 0 of (ln(x^2+1))/x is 0

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limit as x approaches 0 of (ln(x^2+1))/x

x \to 0 lim (\frac{ln ( x ^{2} + 1 )}{x})

0

Solution steps

x \to 0 lim (\frac{ln ( x ^{2} + 1 )}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{\frac{2 x}{x ^{2} + 1}}{1})

= x \to 0 lim (\frac{\frac{2 x}{x ^{2} + 1}}{1})

Plug in the value

x = 0

= \frac{\frac{2 \cdot 0}{0 ^{2} + 1}}{1}

Simplify

\frac{\frac{2 \cdot 0}{0 ^{2} + 1}}{1} : 0

= 0

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What is the limit as x approaches 0 of (ln(x^2+1))/x ?
The limit as x approaches 0 of (ln(x^2+1))/x is 0