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

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

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limit as x approaches 0 of (2^x-5^x)/x

x \to 0 lim (\frac{2 ^{x} - 5 ^{x}}{x})

ln (2) - ln (5)

Decimal Notation

- 0.91629 \dots

Solution steps

x \to 0 lim (\frac{2 ^{x} - 5 ^{x}}{x})

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= ln (2) - ln (5)

\frac{d}{d x} (2 x^{3} - 3 x^{2})

\int (\frac{9}{x} + \frac{x}{9}) d x

y = - 2 x + 2

\int (- 20 x^{3} - 6 x^{- 3}) d x

\frac{\partial}{\partial x} (\frac{7 y - 7 x}{y ^{2} + 2 x ^{2}})

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