What is the limit as x approaches 0 of (7^x-3^x)/x ?

The limit as x approaches 0 of (7^x-3^x)/x is ln(7)-ln(3)

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limit as x approaches 0 of (7^x-3^x)/x

x \to 0 lim (\frac{7 ^{x} - 3 ^{x}}{x})

ln (7) - ln (3)

Decimal

0.84729 \dots

Solution steps

x \to 0 lim (\frac{7 ^{x} - 3 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{7 ^{x} ln ( 7 ) - 3 ^{x} ln ( 3 )}{1})

= x \to 0 lim (\frac{7 ^{x} ln ( 7 ) - 3 ^{x} ln ( 3 )}{1})

Plug in the value

x = 0

= \frac{7 ^{0} ln ( 7 ) - 3 ^{0} ln ( 3 )}{1}

Simplify

\frac{7 ^{0} ln ( 7 ) - 3 ^{0} ln ( 3 )}{1} : ln (7) - ln (3)

= ln (7) - ln (3)

x \to 3 lim (\frac{1}{2 x})

x \to 0 + lim ((1 + x^{3})^{\frac{2}{x}})

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

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

x \to \frac{π}{2} lim (sin (\frac{π}{2}))

What is the limit as x approaches 0 of (7^x-3^x)/x ?
The limit as x approaches 0 of (7^x-3^x)/x is ln(7)-ln(3)