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

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

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limit as x approaches 0 of (4^x-2^x)/x

x \to 0 lim (\frac{4 ^{x} - 2 ^{x}}{x})

ln (2)

Solution steps

x \to 0 lim (\frac{4 ^{x} - 2 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= ln (2)

\frac{d}{d x} (a - b x^{2})

\int_{1}^{2} (2 x - 3)^{2} d x

y^{^{'}} + 7 y = t + e^{- 4 t}

\frac{d}{d x} (4 (3 - 9 x)^{3})

\frac{2}{3 s ^{2} + 6 s + 2}

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