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

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

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

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

- \frac{1}{2}

Decimal

- 0.5

Solution steps

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

Simplify

\frac{1}{x ^{2}} - \frac{1}{ln ( 1 + x ^{2} )} : \frac{ln ( x ^{2} + 1 ) - x ^{2}}{x ^{2} ln ( x ^{2} + 1 )}

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

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- \frac{2 x ^{3}}{x ^{2} + 1}}{2 x ln ( x ^{2} + 1 ) + \frac{2 x ^{3}}{x ^{2} + 1}})

= x \to 0 lim (\frac{- \frac{2 x ^{3}}{x ^{2} + 1}}{2 x ln ( x ^{2} + 1 ) + \frac{2 x ^{3}}{x ^{2} + 1}})

Simplify

\frac{- \frac{2 x ^{3}}{x ^{2} + 1}}{2 x ln ( x ^{2} + 1 ) + \frac{2 x ^{3}}{x ^{2} + 1}} : - \frac{x ^{2}}{ln ( x ^{2} + 1 ) ( x ^{2} + 1 ) + x ^{2}}

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

Apply L'Hopital's Rule

: x \to 0 lim (\frac{- 2 x}{4 x + 2 x ln ( x ^{2} + 1 )})

= x \to 0 lim (\frac{- 2 x}{4 x + 2 x ln ( x ^{2} + 1 )})

Simplify

\frac{- 2 x}{4 x + 2 x ln ( x ^{2} + 1 )} : - \frac{1}{2 + ln ( x ^{2} + 1 )}

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

Plug in the value

x = 0

= - \frac{1}{2 + ln ( 0 ^{2} + 1 )}

Simplify

- \frac{1}{2 + ln ( 0 ^{2} + 1 )} : - \frac{1}{2}

= - \frac{1}{2}

x \to 4 - lim (\frac{a x + b}{x + c})

x \to \infty lim (x^{2} + 3 x + 19 - x)

x \to 0 lim (x^{3} + x^{2} + 3)

x \to \frac{π}{6} lim (sec^{2} (x))

x \to \infty lim (x \cdot 10)

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