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

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

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

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

ln (5) - ln (2)

Decimal

0.91629 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= ln (5) - ln (2)

x \to \infty lim (- a x)

x \to 0 lim (\frac{6 ^{x} - 3 ^{x}}{x})

x \to - \infty lim (- e^{2 x})

x \to 1 + lim (x^{\frac{9}{x - 1}})

t \to 5 lim (g (t))

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