Limits without L'Hopital $\lim\limits_{x\to 0} \frac{e^{3x}-e^{-4x}}{\sin5x}.$

Question

I have a problem finding

$$\lim_{x\to 0} \frac{e^{3x}-e^{-4x}}{\sin5x}.$$

I know that L'hopital's rule can be used to get the answer of $\dfrac{7}{5}$, but I tried to do it without the L'hopital. I tried to solved it by using series expansion and got the correct answer, but the method took so much time.

I want to know that if there is any other method to solve the problem?

Answer 1

By standard limits, note that

$$\frac{e^{3x}-e^{-4x}}{\sin5x}=e^{-4x}\frac{e^{7x}-1}{7x}\frac{5x}{\sin5x}\frac{7x}{5x}\to 1\cdot 1\cdot 1\cdot\frac75=\frac75$$

Answer 2

I do not see how it takes so much time to write the series expansion. If you limit yourself to degree $1$ in $x$ it is quite a one-liner.

$\dfrac{e^{3x}-e^{-4x}}{\sin(5x)}=\dfrac{(1+3x+o(x))-(1-4x+o(x))}{5x+o(x)}=\dfrac{7x+o(x)}{5x+o(x)}\to \dfrac 75$

Answer 3

$$\dfrac{e^{3x}-e^{-4x}}{\sin5x}=\dfrac75e^{-4x}\cdot\dfrac{e^{7x}-1}{7x}\cdot\dfrac{5x}{\sin5x}$$

Answer 4

I dont know how you have proceeded with taylor expansion because

$$ \frac{e^{3x}-e^{-4x}}{\sin\left(5x\right)}\underset{(0)}{\sim}\frac{e^{3x}\left(1-e^{-7x}\right)}{5x}\underset{(0)}{\sim}\frac{1-e^{-7x}}{5x}\underset{(0)}{\sim}-\frac{-7x}{5x}=\frac{7}{5} $$

It was not that long, but maybe it it easier if you factorize by $e^{3x}$.

Answer 5

\begin{align*} \lim_{x \to 0}\frac{e^{3x}-e^{-4x}}{\sin 5x}&=\lim_{x\to 0}\left(\frac{e^{3x}-e^{-4x}}{x}\cdot\frac{x}{\sin 5x}\right)\\[5pt] &=\lim_{x\to 0}\frac{e^{-4x}\left(e^{7x}-1\right)}{x}\cdot\lim_{x\to 0}\frac{x}{\sin 5x}\\[5pt] &=\left(\lim_{x\to 0} e^{-4x}\right)\left(\lim_{x\to 0}\frac{e^{7x}-1}{x}\right)\left(\lim_{x\to 0}\frac{5x}{5\sin 5x}\right)\\[5pt] &=1\cdot\left(\left.\frac{d\,e^{7x}}{dx}\right|_{x=0}\right)\cdot\frac1{5\lim_{x\to 0}\frac{\sin 5x}{5x}}\\[5pt] &=1\cdot 7\cdot\frac{1}{5}\\[5pt] &=\color{red}{\frac75} \end{align*}

Answer 6

Knowing from classical limits that $$\frac{e^{h}-1}{h}=\lim_{u\to0} \frac{u}{\sin u}=1$$ we have $$\lim_{x\to0}\frac{e^{3x}-e^{-4x}}{\sin5x}=\lim_{x\to0}e^{-4x}\frac{e^{7x}-1}{7x}\frac{5x}{\sin5x}\frac{7x}{5x}= \frac75\lim_{x\to0}e^{-4x} \cdot \lim_{h\to0} \frac{e^{h}-1}{h}\cdot\lim_{u\to0} \frac{u}{\sin u}= \frac75$$

Answer 7

Since no one seems to have used "the" definition of hyperbolic sine here... (Actually, there are a variety of definitions. Its power series is probably the easiest way to get the limit used here, just as it is for sine.) \begin{align*} \lim_{x \rightarrow 0} \frac{\mathrm{e}^{3x} - \mathrm{e}^{-4x}}{\sin 5x} &= \lim_{x \rightarrow 0} \frac{\mathrm{e}^{7x/2} - \mathrm{e}^{-7x/2}}{2} \cdot 2\mathrm{e}^{-x/2} \cdot \frac{5x}{\sin 5x}\cdot \frac{1}{5x} \\ &= \lim_{x \rightarrow 0} \sinh(7x/2) \cdot 2\mathrm{e}^{-x/2} \cdot \frac{5x}{\sin 5x}\cdot \frac{1}{5x} \\ &= \lim_{x \rightarrow 0} \frac{\sinh(7x/2)}{7x/2} \cdot 7x\mathrm{e}^{-x/2} \cdot \frac{5x}{\sin 5x}\cdot \frac{1}{5x} \\ &= \lim_{x \rightarrow 0} \frac{\sinh(7x/2)}{7x/2} \cdot \mathrm{e}^{-x/2} \cdot \frac{5x}{\sin 5x}\cdot \frac{7x}{5x} \\ &= 1 \cdot 1 \cdot 1 \cdot \frac{7}{5} \text{.} \end{align*}