What is the limit as x approaches 0 of ((2^x-7^x))/x ?

The limit as x approaches 0 of ((2^x-7^x))/x is ln(2)-ln(7)

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limit as x approaches 0 of ((2^x-7^x))/x

x \to 0 lim (\frac{( 2 ^{x} - 7 ^{x} )}{x})

ln (2) - ln (7)

Decimal Notation

- 1.25276 \dots

Solution steps

x \to 0 lim (\frac{2 ^{x} - 7 ^{x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 ^{x} ln ( 2 ) - 7 ^{x} ln ( 7 )}{1})

= x \to 0 lim (\frac{2 ^{x} ln ( 2 ) - 7 ^{x} ln ( 7 )}{1})

Plug in the value

x = 0

= \frac{2 ^{0} ln ( 2 ) - 7 ^{0} ln ( 7 )}{1}

Simplify

\frac{2 ^{0} ln ( 2 ) - 7 ^{0} ln ( 7 )}{1} : ln (2) - ln (7)

= ln (2) - ln (7)

\frac{1}{s ( s ^{2} + 2 s + 5 )}

\frac{1}{y ^{2}} - \frac{3}{y ^{4}}

\frac{d y}{d x} = e^{x - y}

f (x) = x^{3} ln (2 x), at x = 1

\int_{1}^{9} y ln (y) d y

What is the limit as x approaches 0 of ((2^x-7^x))/x ?
The limit as x approaches 0 of ((2^x-7^x))/x is ln(2)-ln(7)