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

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

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limit as x approaches+0 of (2^x-3^x)/x

x \to + 0 lim (\frac{2 ^{x} - 3 ^{x}}{x})

ln (2) - ln (3)

Decimal Notation

- 0.40546 \dots

Solution steps

x \to 0 lim (\frac{2 ^{x} - 3 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= ln (2) - ln (3)

e^{9 e^{x}}

\frac{d y}{d x} = \frac{1}{6 x - 42}

y = \frac{e ^{4 x^{2} + 1}}{8}

2 x + 5 x

h (x) = x^{2}

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