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

The limit as x approaches 0 of (e^x-2^x)/x is 1-ln(2)

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

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

1 - ln (2)

Decimal Notation

0.30685 \dots

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= 1 - ln (2)

f (x) = ln (4 - x^{2} + 2 x^{4}), at x = 1

f (t) = \frac{9}{4} t^{2} \cdot e^{2 t} \cdot e^{3 t}

\int 128 cos^{4} (8 x) d x

\int y^{2} y d y

\frac{d}{d x} (\frac{8 x ^{2} + 2 x + 6}{x})

What is the limit as x approaches 0 of (e^x-2^x)/x ?
The limit as x approaches 0 of (e^x-2^x)/x is 1-ln(2)