What is the limit as x approaches 0+of 7xln(x) ?

The limit as x approaches 0+of 7xln(x) is 0

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limit as x approaches 0+of 7xln(x)

x \to 0 + lim (7 x ln (x))

0

Solution steps

x \to 0 + lim (7 x ln (x))

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

= 7 \cdot x \to 0 + lim (x ln (x))

Rewrite for L'Hopital

= 7 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{x}})

Apply L'Hopital's Rule

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

= 7 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- \frac{1}{x ^{2}}})

Simplify

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

= 7 \cdot x \to 0 + lim (- x)

Plug in the value

x = 0

= 7 (- 0)

Simplify

= 0

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

x \to 0 lim (x^{2} sin (\frac{π}{x}))

x \to 3 lim (x - 7)

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

z \to i lim (\frac{z ^{4} - 1}{z - i})

What is the limit as x approaches 0+of 7xln(x) ?
The limit as x approaches 0+of 7xln(x) is 0