What is the limit as x approaches 0 of (2^{3+x}-8)/x ?

The limit as x approaches 0 of (2^{3+x}-8)/x is 8ln(2)

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limit as x approaches 0 of (2^{3+x}-8)/x

x \to 0 lim (\frac{2 ^{3 + x} - 8}{x})

8 ln (2)

Decimal Notation

5.54517 \dots

Solution steps

x \to 0 lim (\frac{2 ^{3 + x} - 8}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{ln ( 2 ) \cdot 2 ^{x + 3}}{1})

= x \to 0 lim (\frac{ln ( 2 ) \cdot 2 ^{x + 3}}{1})

Plug in the value

x = 0

= \frac{ln ( 2 ) \cdot 2 ^{0 + 3}}{1}

Simplify

= 8 ln (2)

\int x^{2} (3 + x^{3})^{3} d x

\frac{d}{d x} (5 + 7 x)

\int \frac{1}{( 9 x ^{2} + 16 ) ^{2}} d x

\frac{d}{d x} (2 x (12 - x)^{3})

ln (1 + 8 x^{2})

What is the limit as x approaches 0 of (2^{3+x}-8)/x ?
The limit as x approaches 0 of (2^{3+x}-8)/x is 8ln(2)