What is the limit as x approaches+0 of (e^{2x}-1)/(sin(3x)) ?

The limit as x approaches+0 of (e^{2x}-1)/(sin(3x)) is 2/3

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limit as x approaches+0 of (e^{2x}-1)/(sin(3x))

x \to + 0 lim (\frac{e ^{2 x} - 1}{sin ( 3 x )})

\frac{2}{3}

Solution steps

x \to 0 lim (\frac{e ^{2 x} - 1}{sin ( 3 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 e ^{2 x}}{cos ( 3 x ) \cdot 3})

= x \to 0 lim (\frac{2 e ^{2 x}}{cos ( 3 x ) \cdot 3})

Plug in the value

x = 0

= \frac{2 e ^{2 \cdot 0}}{cos ( 3 \cdot 0 ) \cdot 3}

Simplify

\frac{2 e ^{2 \cdot 0}}{cos ( 3 \cdot 0 ) \cdot 3} : \frac{2}{3}

= \frac{2}{3}

x \to 3 lim (x^{3} - 2 x^{2} + 1)

n = 1 \sum \infty \frac{30}{3 ^{n}}

y = \frac{2}{1 + x ^{2}}, y = ∣ x ∣

\int \frac{2 x}{4 - x ^{2}} d x

\int 4 e^{7 x} d x

What is the limit as x approaches+0 of (e^{2x}-1)/(sin(3x)) ?
The limit as x approaches+0 of (e^{2x}-1)/(sin(3x)) is 2/3