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)

Decimal Notation

0.69314 \dots

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)

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

\int (sin (x) + 7 cos (x) - 1) d x

(t e)^{3} e^{- 2 (t + 1)}

n = 5 \sum \infty e^{5 - 2 n}

\frac{3}{s ^{2} - 2 s - 15}

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)