What is the limit as n approaches 0+of n^2ln(4n^2) ?

The limit as n approaches 0+of n^2ln(4n^2) is 0

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limit as n approaches 0+of n^2ln(4n^2)

n \to 0 + lim (n^{2} ln (4 n^{2}))

0

Solution steps

n \to 0 + lim (n^{2} ln (4 n^{2}))

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: n \to 0 + lim (\frac{\frac{2}{n}}{- \frac{2}{n ^{3}}})

= n \to 0 + lim (\frac{\frac{2}{n}}{- \frac{2}{n ^{3}}})

Simplify

\frac{\frac{2}{n}}{- \frac{2}{n ^{3}}} : - n^{2}

= n \to 0 + lim (- n^{2})

Plug in the value

n = 0

= - 0^{2}

Simplify

= 0

x \to 2 π lim (2 x cos (x))

x \to 0 lim (175)

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

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

x \to - \frac{3}{2} - lim (e^{- 0.2 x})

What is the limit as n approaches 0+of n^2ln(4n^2) ?
The limit as n approaches 0+of n^2ln(4n^2) is 0