What is the limit as x approaches 0 of (9^x-5^x)/x ?

The limit as x approaches 0 of (9^x-5^x)/x is 2ln(3)-ln(5)

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limit as x approaches 0 of (9^x-5^x)/x

x \to 0 lim (\frac{9 ^{x} - 5 ^{x}}{x})

2 ln (3) - ln (5)

Decimal Notation

0.58778 \dots

Solution steps

x \to 0 lim (\frac{9 ^{x} - 5 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 ln ( 3 ) \cdot 9 ^{x} - 5 ^{x} ln ( 5 )}{1})

= x \to 0 lim (\frac{2 ln ( 3 ) \cdot 9 ^{x} - 5 ^{x} ln ( 5 )}{1})

Plug in the value

x = 0

= \frac{2 ln ( 3 ) \cdot 9 ^{0} - 5 ^{0} ln ( 5 )}{1}

Simplify

\frac{2 ln ( 3 ) \cdot 9 ^{0} - 5 ^{0} ln ( 5 )}{1} : 2 ln (3) - ln (5)

= 2 ln (3) - ln (5)

\frac{\partial}{\partial y} (\frac{y ln ( y )}{x - 1})

\int \frac{2}{x 9 + x ^{2}} d x

\int \frac{6 x + 5}{( 3 x ^{2} + 5 x ) ^{6}} d x

f (x) = \frac{( 1 - 4 x )}{3 + x}

\frac{d}{d x} (1 0^{9 - x^{2}})

What is the limit as x approaches 0 of (9^x-5^x)/x ?
The limit as x approaches 0 of (9^x-5^x)/x is 2ln(3)-ln(5)