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

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

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limit as t approaches 0+of t(ln(t))^2

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

= t \to 0 + lim (\frac{\frac{2 l n ( t )}{t}}{- \frac{1}{t ^{2}}})

Simplify

\frac{\frac{2 l n ( t )}{t}}{- \frac{1}{t ^{2}}} : - 2 t ln (t)

= t \to 0 + lim (- 2 t ln (t))

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

= - 2 \cdot t \to 0 + lim (t ln (t))

Rewrite for L'Hopital

= - 2 \cdot t \to 0 + lim (\frac{ln ( t )}{\frac{1}{t}})

Apply L'Hopital's Rule

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

Simplify

\frac{\frac{1}{t}}{- \frac{1}{t ^{2}}} : - t

= - 2 \cdot t \to 0 + lim (- t)

Plug in the value

t = 0

= - 2 (- 0)

Simplify

= 0

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