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

- 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})

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)

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)