What is the limit as x approaches 0 of (e^{3x}-e^{-3x})/(4x) ?

The limit as x approaches 0 of (e^{3x}-e^{-3x})/(4x) is 3/2

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limit as x approaches 0 of (e^{3x}-e^{-3x})/(4x)

x \to 0 lim (\frac{e ^{3 x} - e ^{- 3 x}}{4 x})

\frac{3}{2}

Decimal Notation

1.5

Solution steps

x \to 0 lim (\frac{e ^{3 x} - e ^{- 3 x}}{4 x})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{4} \cdot x \to 0 lim (\frac{e ^{3 x} - e ^{- 3 x}}{x})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{3 e ^{3 x} + 3 e ^{- 3 x}}{1})

= \frac{1}{4} \cdot x \to 0 lim (\frac{3 e ^{3 x} + 3 e ^{- 3 x}}{1})

Plug in the value

x = 0

= \frac{1}{4} \cdot \frac{3 e ^{3 \cdot 0} + 3 e ^{- 3 \cdot 0}}{1}

Simplify

\frac{1}{4} \cdot \frac{3 e ^{3 \cdot 0} + 3 e ^{- 3 \cdot 0}}{1} : \frac{3}{2}

= \frac{3}{2}

x^{3} - 15 x^{2} + 72 x + 12

\frac{d}{d x} (\frac{1}{\frac{1}{x}})

\int_{0}^{3} 2 π y (9 - y^{2}) d y

(2, - 2), (9, 3)

y^{(\frac{1}{2})} y^{^{'}} + y^{(\frac{3}{2})} = 1

What is the limit as x approaches 0 of (e^{3x}-e^{-3x})/(4x) ?
The limit as x approaches 0 of (e^{3x}-e^{-3x})/(4x) is 3/2