What is the limit as x approaches 0 of (2^{2x}-1)/x ?

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

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limit as x approaches 0 of (2^{2x}-1)/x

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

2 ln (2)

Decimal

1.38629 \dots

Solution steps

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

2^{2 x} = 4^{x}

= x \to 0 lim (\frac{4 ^{x} - 1}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{ln ( 2 ) \cdot 2 ^{2 x + 1}}{1})

= x \to 0 lim (\frac{ln ( 2 ) \cdot 2 ^{2 x + 1}}{1})

Plug in the value

x = 0

= \frac{ln ( 2 ) \cdot 2 ^{2 \cdot 0 + 1}}{1}

Simplify

= 2 ln (2)

\int ln (z) d z

\int \frac{( x e ^{x} )}{( x + 1 ) ^{2}} d x

\int x \cdot sin (3 x) d x

x cos (π) x, x = 0

f (- 1) = x^{2} + 2 - 1

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