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

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

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

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

2

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 2

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