Find lim as x to 0 (e^(alpha x)- cos alpha x)/(e^(beta

Find $\displaystyle\lim_{x\to0}\frac{e^{\alpha x}-\cos\alpha x}{e^{\beta x}-\cos\beta x}$, where $\beta\neq0$.

Step 1: Check the indeterminate form

At $x = 0$ the numerator is $1-1 = 0$ and the denominator is $1-1 = 0$. The form is $\dfrac00$.

Step 2: Expand using Maclaurin series

Recall

\[ e^{t} = 1+t+\frac{t^{2}}{2}+\cdots, \qquad \cos t = 1-\frac{t^{2}}{2}+\cdots \]

With $t = \alpha x$ the constants cancel and the two quadratic halves add: \[e^{\alpha x}-\cos\alpha x = \left(1+\alpha x+\frac{\alpha^{2}x^{2}}{2}+\cdots\right)-\left(1-\frac{\alpha^{2}x^{2}}{2}+\cdots\right) = \alpha x+\alpha^{2}x^{2}+\cdots \tag{1}\]

The same computation with $t = \beta x$ gives

\[ e^{\beta x}-\cos\beta x = \beta x+\beta^{2}x^{2}+\cdots \tag{2} \]

Step 3: Form the ratio

For $x\neq0$, divide (1) by (2) and cancel the common factor $x$: \[\frac{e^{\alpha x}-\cos\alpha x}{e^{\beta x}-\cos\beta x} = \frac{\alpha x+\alpha^{2}x^{2}+\cdots}{\beta x+\beta^{2}x^{2}+\cdots} = \frac{x\bigl(\alpha+\alpha^{2}x+\cdots\bigr)}{x\bigl(\beta+\beta^{2}x+\cdots\bigr)} = \frac{\alpha+\alpha^{2}x+\cdots}{\beta+\beta^{2}x+\cdots}.\]

The hypothesis $\beta\neq0$ is what makes the new denominator nonzero at the origin.

Step 4: Evaluate the limit

\[ \lim_{x\to0}\frac{\alpha+\alpha^{2}x+\cdots}{\beta+\beta^{2}x+\cdots} = \frac{\alpha}{\beta}. \]

Step 5: Verification

\[\text{By l'H\^opital: } \lim_{x\to0}\frac{\alpha e^{\alpha x}+\alpha\sin\alpha x}{\beta e^{\beta x}+\beta\sin\beta x} = \frac{\alpha}{\beta} \;\;\checkmark\]

\[\alpha = 2,\ \beta = 3:\ \tfrac{\alpha}{\beta} = 0.6666667;\quad x = 0.001:\ 0.6660014;\quad x = -0.01:\ 0.6734812 \;\;\checkmark\]

\[\text{Full series: } e^{\alpha x}-\cos\alpha x = \alpha x+\alpha^{2}x^{2}+\frac{\alpha^{3}x^{3}}{6}+\cdots \;\;\checkmark\]