# Calculate $\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}$

Question:

Calculate

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}$$

using substitution, cancellation, factoring etc. and common standard limits (i.e. not by L'Hôpital's rule).

Attempted solution:

It is not possible to solve by evaluating it directly, since it leads to a "$\frac{0}{0}$" situation. These kinds of problems are typically solved by cancellation and/or direct application of a after some artful substitution in a precalculus context.

It was a bit hard to find a good substitution, so I tried several:

$t = 3x$

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \{t=3x\} = \lim_{x \to 0} \frac{e^{t} - 1}{e^{4\frac{t}{3}} - 1}$$

Although this gets me closer to the standard limit

$$\lim_{x to 0} \frac{e^x - 1}{x} = 1$$

...it is not good enough.

$t = e^{4x} - 1$

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \{t=e^{4x} - 1\} = \lim_{t \to 0} \frac{e^{\frac{3 \ln (t+ 1)}{4}} - 1}{t}$$

Not sure where to move on from here, if it is at all possible.

Looks like the denominator can be factored by the conjugate rule:

$$\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1} = \lim_{x \to 0} \frac{e^{3x} - 1}{(e^{2x} - 1)(e^{2x} + 1)}$$

Unclear where this trail can lead.

What are some productive substitutions or approaches to calculating this limit (without L'Hôpital's rule)?

hint: do a simple trick: $\dfrac{e^{3x}-1}{e^{4x}-1} = \dfrac{e^{3x}-1}{3x}\cdot \dfrac{4x}{e^{4x}-1}\cdot \dfrac{3}{4}$. The first two factors tend to $1$ each, leaving you the last factor being the answer.

• .... which is actually just an application of L'Hospital without mentioning the name. Commented May 25, 2015 at 23:49

Here's another way. $$\frac{e^{3x}-1}{e^{4x}-1}=\frac{(e^x)^3-1}{(e^x)^4-1}=\frac{(e^x-1)(e^{2x}+e^x+1)}{(e^{2x}+1)(e^x+1)(e^x-1)}=\frac{e^{2x}+e^x+1}{(e^{2x}+1)(e^x+1)}.$$

So,

$$\lim_{x\to0}\frac{e^{3x}-1}{e^{4x}-1}=\lim_{x\to0}\frac{e^{2x}+e^x+1}{(e^{2x}+1)(e^x+1)}=\frac{1+1+1}{2\cdot2}=\frac34.$$

• Can you say more? Commented May 25, 2015 at 23:07
• Yep. I fixed it. Commented May 25, 2015 at 23:08

Use the identities $a^3 - b^3 = (a - b) (a^2 + a b + b^2)$ and $a^4 - b^4 = (a - b) (a+b)(a^2 + b^2)$ where $a=e^x$ and $b=1$ to get $\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}= \lim_{x \to 0} \frac{(e^{x} - 1)(e^{2x} +e^{x}+ 1)}{(e^{x} - 1)(e^{x} + 1)(e^{2x} + 1)}= \lim_{x \to 0} \frac{(e^{2x} +e^{x}+ 1)}{(e^{x} + 1)(e^{2x} + 1)}=\frac{3}{4}$

Hint: $$\frac{e^{3x}-1}{e^{4x}-1} = \frac{e^{3x}-1}{3x}\cdot\frac{4x}{e^{4x}-1}\cdot \frac{3}{4}$$

For positive integers $m$ and $n$,

$\begin{array}\\ \lim_{x \to 0} \frac{e^{mx}-1}{e^{nx}-1} &=\lim_{x \to 0} \frac{(e^x-1)(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)}{(e^x-1)(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)}\\ &=\lim_{x \to 0} \frac{e^{(m-1)x}+e^{(m-2)x}+...+e^x+1}{e^{(m-1)x}+e^{(m-2)x}+...+e^x+1} \quad\text{(true for$x \ne 0)$}\\ &= \frac{\lim_{x \to 0}(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)}{\lim_{x \to 0}(e^{(m-1)x}+e^{(m-2)x}+...+e^x+1)} \quad\text{(since the numerator and denominator are$> 1)$}\\ &=\frac{m}{n} \end{array}$

Then put $m=3$ and $n=4$.

$$\frac{(e^x)^3-1}{(e^x)^4-1}=\frac{e^x+1}{2(e^{2x}+1)}+\frac{1}{2(e^x+1)}$$

Hint: You presumably already know how to prove L'hospital's Rule in general. Try adapting that general proof to this specific case.

You can use l'Hopital's Rule to get $\lim_{x \to 0} \frac{e^{3x} - 1}{e^{4x} - 1}= \lim_{x \to 0} \frac{3e^{3x}}{4e^{4x} }= \frac{3}{4}.$