How to find the Value of the Limit $ \lim _{x\to 0}\left(\dfrac{\sin\alpha x-\sin\beta x}{e^{\alpha x}-e^{\beta x}}\right)$

Question

I am familiar with indeterminate limits which can be solved using L'Hospital Rule, but I cannot use that in this case.

Answer 1

Using the rules of L'Hospital we get $$\lim_{x\to 0}\frac{\alpha\cos(\alpha x)-\beta\cos(\beta x)}{\alpha e^{\alpha x}-\beta e^{\beta x}}$$ Can you finish?

Answer 2

If you are against using L'Hospital, you may just expand the functions.

In the limit $\eta x\to0$ $$ \sin(\eta x) \approx \eta x + O(x^3)\\ \exp(\eta x) \approx 1 + \eta x + O(x^2) $$ You should be able to continue from here ...

Answer 3

W/O L'Hospital and Series Expansion,

$$\lim_{x\to0}\dfrac{\sin2ax-\sin2bx}{e^{2ax}-e^{2bx}}$$

$$=\lim_{x\to0}\dfrac{2\sin(a-b)x}{e^{2(a-b)x}-1}\cdot\lim_{x\to0}\dfrac{\cos(a+b)x}{e^{2bx}}$$

$$=\dfrac{\lim_{x\to0}\dfrac{\sin(a-b)x}{(a-b)x}}{\lim_{x\to0}\dfrac{e^{2(a-b)x}-1}{2(a-b)x}}=?$$

Answer 4

Hint: Suppose that $\alpha \ne \beta$. Note that $\sin$ and $\exp$ are differentiable on $\mathbb{R}$, and we can use the mean value theorem. There are $\gamma_1$, $\gamma_2$ in $(\alpha x,\beta x)\cup (\beta x,\alpha x)$ such that $$ \sin \alpha x - \sin\beta x=(\alpha x-\beta x)\cos \gamma_1,\qquad e^{\alpha x}-e^{\beta x}=(\alpha x-\beta x)e^{\gamma_2}. $$

Answer 5

Assume that you know $\sin{x}\sim x$ and $e^x \sim x+1$ both as $x\to0$. Then $$\lim_{x\to0}\dfrac{\sin\alpha x-\sin\beta x}{e^{\alpha x}-e^{\beta x}}=\lim_{x\to0}\dfrac{\sin\beta x(\dfrac{\sin{\alpha x}}{\sin{\beta x}}-1)}{(e^{\beta x}-1)(\dfrac{e^{\alpha x}-1}{e^{\beta x}-1}-1)}$$

Answer 6

By Cauchy's Mean Value Theorem we see that the expression under limit is equal to $\dfrac{\cos c} {e^c} $ where $c$ lies between $\alpha x$ and $\beta x$. As $x\to 0$ we have $c\to 0$ and the desired limit is equal to $1$.