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

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

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limit as x approaches 0 of ((e^{ax}-e^{bx}))/(sin(ax)-sin(bx))

x \to 0 lim (\frac{( e ^{a x} - e ^{b x} )}{sin ( a x ) - sin ( b x )})

1

Solution steps

x \to 0 lim (\frac{e ^{a x} - e ^{b x}}{sin ( a x ) - sin ( b x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{e ^{a x} a - e ^{b x} b}{cos ( a x ) a - cos ( b x ) b})

= x \to 0 lim (\frac{e ^{a x} a - e ^{b x} b}{cos ( a x ) a - cos ( b x ) b})

Plug in the value

x = 0

= \frac{e ^{a \cdot 0} a - e ^{b \cdot 0} b}{cos ( a \cdot 0 ) a - cos ( b \cdot 0 ) b}

Simplify

\frac{e ^{a \cdot 0} a - e ^{b \cdot 0} b}{cos ( a \cdot 0 ) a - cos ( b \cdot 0 ) b} : 1

= 1

x \to - 2 - lim (3 x + 1)

x \to \infty lim (\frac{6 - 9 x + 3 x ^{2}}{- x ^{2} + 4 x - 3})

x \to \infty lim (arctanh (e^{2 x}))

x \to - 4 + lim (\frac{x}{x ^{2} - 16})

t \to 1 lim (\frac{tan ( t )}{t})

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