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

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

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

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

ln (7) - ln (2)

Decimal

1.25276 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= ln (7) - ln (2)

\frac{d y}{d x} + 2 x y + x y^{4} = 0

y^{^{'}} + (6 x + \frac{1}{x}) y = 1

x \frac{d y}{d x} + 4 y = 5

\int 6 x e^{3 x^{2}} d x

a y^{^{''}} + b y^{^{'}} = 0

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