What is the limit as x approaches+0 of (5^x-7^x)/x ?

The limit as x approaches+0 of (5^x-7^x)/x is ln(5)-ln(7)

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limit as x approaches+0 of (5^x-7^x)/x

x \to + 0 lim (\frac{5 ^{x} - 7 ^{x}}{x})

ln (5) - ln (7)

Decimal Notation

- 0.33647 \dots

Solution steps

x \to 0 lim (\frac{5 ^{x} - 7 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{5 ^{x} ln ( 5 ) - 7 ^{x} ln ( 7 )}{1})

= x \to 0 lim (\frac{5 ^{x} ln ( 5 ) - 7 ^{x} ln ( 7 )}{1})

Plug in the value

x = 0

= \frac{5 ^{0} ln ( 5 ) - 7 ^{0} ln ( 7 )}{1}

Simplify

\frac{5 ^{0} ln ( 5 ) - 7 ^{0} ln ( 7 )}{1} : ln (5) - ln (7)

= ln (5) - ln (7)

y^{^{'}} = - a \cdot y + b \cdot y

f (x) = x^{2} + 2 (3.11)

- 4 cos (6 t)

\int (5^{x}) d x

\frac{d}{d x} (5 sin (y) + 3 e^{2 x})

What is the limit as x approaches+0 of (5^x-7^x)/x ?
The limit as x approaches+0 of (5^x-7^x)/x is ln(5)-ln(7)