Find lim as x to 0 (e^2x-e^bx)/(sin x)

Find $\displaystyle\lim_{x\to0}\frac{e^{2x}-e^{bx}}{\sin x}$.

Step 1: The form

At $x = 0$ the numerator is $e^{0}-e^{0} = 0$ and $\sin0 = 0$: the form is $\dfrac00$.

Step 2: Apply l'Hôpital's rule

\[ \frac{d}{dx}\bigl(e^{2x}-e^{bx}\bigr) = 2e^{2x}-b\,e^{bx}, \qquad \frac{d}{dx}\sin x = \cos x, \]

\[ \lim_{x\to0}\frac{e^{2x}-e^{bx}}{\sin x} = \lim_{x\to0}\frac{2e^{2x}-b\,e^{bx}}{\cos x} = \frac{2-b}{1}. \]

Step 3: The same result from the series

Expanding each exponential to first order, $e^{2x} = 1+2x+O(x^{2})$ and $e^{bx} = 1+bx+O(x^{2})$, so the constants cancel: \[\frac{e^{2x}-e^{bx}}{\sin x} = \frac{\bigl(2-b\bigr)x+O(x^{2})}{x+O(x^{3})} \longrightarrow 2-b.\]

Step 4: Verification

\[b = 3 \Rightarrow 2-b = -1:\quad x = 0.01:\ -1.0253365; \qquad x = -0.01:\ -0.9753302 \;\;\checkmark\]

\[x = 0.001:\ -1.0025033 \;\;\checkmark\]

\[b = 2:\ \text{the numerator is identically } 0 \text{ and the limit is } 0 \;\;\checkmark\]