What is the limit as x approaches 0 of ((6^x-2^x))/x ?

The limit as x approaches 0 of ((6^x-2^x))/x is ln(6)-ln(2)

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limit as x approaches 0 of ((6^x-2^x))/x

x \to 0 lim (\frac{( 6 ^{x} - 2 ^{x} )}{x})

ln (6) - ln (2)

Decimal

1.09861 \dots

Solution steps

x \to 0 lim (\frac{6 ^{x} - 2 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - 2 ^{x} ln ( 2 )}{1})

= x \to 0 lim (\frac{6 ^{x} ln ( 6 ) - 2 ^{x} ln ( 2 )}{1})

Plug in the value

x = 0

= \frac{6 ^{0} ln ( 6 ) - 2 ^{0} ln ( 2 )}{1}

Simplify

\frac{6 ^{0} ln ( 6 ) - 2 ^{0} ln ( 2 )}{1} : ln (6) - ln (2)

= ln (6) - ln (2)

x \to 0 lim (x^{3} (sin (x) - 1))

x \to \frac{1}{3} lim (\frac{arcsin ( x ) - arcsin ( \frac{1}{3} )}{x - \frac{1}{3}})

x \to 0 + lim (1 + x^{3})

x \to 0 lim (\frac{1}{x ^{1}})

x \to 7 lim (1)

What is the limit as x approaches 0 of ((6^x-2^x))/x ?
The limit as x approaches 0 of ((6^x-2^x))/x is ln(6)-ln(2)