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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Apply L'Hopital's Rule

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

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

Simplify

\frac{\frac{2 x}{1 + x ^{2}}}{2 x} : \frac{1}{1 + x ^{2}}

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

Plug in the value

x = 0

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

Simplify

= \frac{1}{2}

x \to 0 lim (2 e^{x} cos (x))

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

x \to 0 lim (tan (sin (x)))

x \to - 1 lim (x + 6)

y \to - 3 lim (14 y)

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