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 Notation

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 (19)

h \to 0 lim (\frac{ln ( 3 + h ) - ln ( 3 )}{h})

x \to 0 + lim (\frac{ln ( 1 + x )}{x})

x \to 0 + lim (1^{\frac{l n ( 6 )}{x}})

x \to 0 lim ((1 + 8 x)^{\frac{1}{x}})

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)