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

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

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limit as x approaches 0+of sqrt(x)ln(7x)

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Simplify

\frac{\frac{1}{x}}{- \frac{1}{2 x ^{\frac{3}{2}}}} : - 2 x^{\frac{1}{2}}

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

Plug in the value

x = 0

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

Simplify

= 0

x \to 0 lim (- 9)

x \to 2 - lim (\frac{x ^{2} + 5}{x + 1})

x \to 8 + lim (\frac{1}{8 - x})

x \to 2 lim (5^{x})

x \to h lim (sin (x))

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