What is the limit as x approaches 0 of (x4^x)/(4^x-1) ?

The limit as x approaches 0 of (x4^x)/(4^x-1) is 1/(2ln(2))

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limit as x approaches 0 of (x4^x)/(4^x-1)

x \to 0 lim (\frac{x 4 ^{x}}{4 ^{x} - 1})

\frac{1}{2 ln ( 2 )}

Decimal

0.72134 \dots

Solution steps

x \to 0 lim (\frac{x \cdot 4 ^{x}}{4 ^{x} - 1})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{4 ^{0} + ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1} \cdot 0}{ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}

Simplify

\frac{4 ^{0} + ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1} \cdot 0}{ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}} : \frac{1}{2 ln ( 2 )}

= \frac{1}{2 ln ( 2 )}

\int (π sin (π) x) d x

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

4 x^{2} + y = 4, x^{4} - y = 1

x \to \infty + lim (\frac{x ^{5} - 3 x ^{2} + 5 x - 9}{7 x ^{4} - 6 x ^{2} + 5 x - 9})

\frac{d}{d x} (e^{l n (3)})

What is the limit as x approaches 0 of (x4^x)/(4^x-1) ?
The limit as x approaches 0 of (x4^x)/(4^x-1) is 1/(2ln(2))