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

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

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

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

0

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 0

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