What is the limit as x approaches 0 of (5^x-4^x)/x ?

The limit as x approaches 0 of (5^x-4^x)/x is ln(5)-2ln(2)

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limit as x approaches 0 of (5^x-4^x)/x

x \to 0 lim (\frac{5 ^{x} - 4 ^{x}}{x})

ln (5) - 2 ln (2)

Decimal Notation

0.22314 \dots

Solution steps

x \to 0 lim (\frac{5 ^{x} - 4 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{5 ^{x} ln ( 5 ) - ln ( 2 ) \cdot 2 ^{2 x + 1}}{1})

= x \to 0 lim (\frac{5 ^{x} ln ( 5 ) - ln ( 2 ) \cdot 2 ^{2 x + 1}}{1})

Plug in the value

x = 0

= \frac{5 ^{0} ln ( 5 ) - ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}{1}

Simplify

\frac{5 ^{0} ln ( 5 ) - ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}{1} : ln (5) - 2 ln (2)

= ln (5) - 2 ln (2)

\frac{2 x + 3}{3 x + 2}

\int_{0}^{1} \frac{z}{( z ^{2} + 1 ) ^{3}} d z

x \to 0 + lim (ln (x^{3}))

(1 + t^{2}) y^{^{'}} + 4 t y = t

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

What is the limit as x approaches 0 of (5^x-4^x)/x ?
The limit as x approaches 0 of (5^x-4^x)/x is ln(5)-2ln(2)