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 Notation

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} ln (x))

f (x) = \frac{x}{x ^{2} + 9}

x (x + y) y^{^{'}} + y (3 x + y) = 0

\int 5 cos (3 z) + \frac{2}{e ^{4 z}} d z

2 t y^{^{'}} + y = e^{t}

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)