What is the limit as x approaches 0 of (6^x-3^x)/x ?

The limit as x approaches 0 of (6^x-3^x)/x is ln(6)-ln(3)

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limit as x approaches 0 of (6^x-3^x)/x

x \to 0 lim (\frac{6 ^{x} - 3 ^{x}}{x})

ln (6) - ln (3)

Decimal

0.69314 \dots

Solution steps

x \to 0 lim (\frac{6 ^{x} - 3 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - 3 ^{x} ln ( 3 )}{1})

= x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - 3 ^{x} ln ( 3 )}{1})

Plug in the value

x = 0

= \frac{6 ^{0} ln ( 6 ) - 3 ^{0} ln ( 3 )}{1}

Simplify

\frac{6 ^{0} ln ( 6 ) - 3 ^{0} ln ( 3 )}{1} : ln (6) - ln (3)

= ln (6) - ln (3)

x \to - \infty lim (- e^{2 x})

x \to 1 + lim (x^{\frac{9}{x - 1}})

t \to 5 lim (g (t))

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

x \to 3 - lim (\frac{5}{3 - x})

What is the limit as x approaches 0 of (6^x-3^x)/x ?
The limit as x approaches 0 of (6^x-3^x)/x is ln(6)-ln(3)