What is the limit as x approaches 0 of (3^x-4^x)/x ?

The limit as x approaches 0 of (3^x-4^x)/x is ln(3)-2ln(2)

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limit as x approaches 0 of (3^x-4^x)/x

x \to 0 lim (\frac{3 ^{x} - 4 ^{x}}{x})

ln (3) - 2 ln (2)

Decimal

- 0.28768 \dots

Solution steps

x \to 0 lim (\frac{3 ^{x} - 4 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{3 ^{0} ln ( 3 ) - ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}{1}

Simplify

\frac{3 ^{0} ln ( 3 ) - ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}{1} : ln (3) - 2 ln (2)

= ln (3) - 2 ln (2)

n = 17 \sum \infty \frac{n + 8}{n !}

\int - x^{2} e^{2 x} d x

\frac{d}{d x} (\frac{1}{2} tan (x) sin (2 x))

\int 25 sec^{5} (x) d x

y^{^{''}} + 7 y^{^{'}} + 10 y = 14 x e^{3 x}

What is the limit as x approaches 0 of (3^x-4^x)/x ?
The limit as x approaches 0 of (3^x-4^x)/x is ln(3)-2ln(2)