Evaluate the following limits without using L'Hôpital's Rule: lim x-0 (6e3x - 6)/(e^x - 1) and lim x-0 (e4x…

Question

Evaluate the following limits without using L'Hôpital's Rule: lim x->0 (6e3x - 6)/(e^x - 1) and lim x->0 (e4x - e2x)/x.

ScanSolve AI Tutor · Accepted Answer

Okay, ~Mataa🫶🏽, let's evaluate these limits without L'Hôpital's Rule.

For the first limit:

\lim_{x \to 0} \frac{6e^{3x} - 6}{e^x - 1}

First, substitute $x=0$ :

\frac{6e^{3 \cdot 0} - 6}{e^0 - 1} = \frac{6 \cdot 1 - 6}{1 - 1} = \frac{0}{0}

This is an indeterminate form. We will use the standard limit $\lim_{x \to 0} \frac{e^{ax} - 1}{x} = a$ .

Factor out 6 from the numerator.
$\lim_{x \to 0} \frac{6(e^{3x} - 1)}{e^x - 1}$
Multiply the numerator and denominator by $x$ to create the standard limit forms.
$\lim_{x \to 0} 6 \cdot \frac{e^{3x} - 1}{x} \cdot \frac{x}{e^x - 1}$
Rearrange the terms to clearly show the standard limits.
$\lim_{x \to 0} 6 \cdot \frac{e^{3x} - 1}{x} \cdot \frac{1}{\frac{e^x - 1}{x}}$
Apply the standard limit $\lim_{x \to 0} \frac{e^{ax} - 1}{x} = a$ .
$6 \cdot \left( \lim_{x \to 0} \frac{e^{3x} - 1}{x} \right) \cdot \left( \lim_{x \to 0} \frac{1}{\frac{e^x - 1}{x}} \right)$ $= 6 \cdot (3) \cdot \frac{1}{1}$ $= 18$
The first limit is:
$\boxed{18}$
For the second limit:
$\lim_{x \to 0} \frac{e^{4x} - e^{2x}}{x}$
First, substitute $x=0$ :
$\frac{e^{4 \cdot 0} - e^{2 \cdot 0}}{0} = \frac{e^0 - e^0}{0} = \frac{1 - 1}{0} = \frac{0}{0}$
This is an indeterminate form. We will again use the standard limit $\lim_{x \to 0} \frac{e^{ax} - 1}{x} = a$ .
Add and subtract 1 in the numerator to create terms that fit the standard limit form.
$\lim_{x \to 0} \frac{e^{4x} - 1 - e^{2x} + 1}{x}$
Group the terms and split the fraction.
$\lim_{x \to 0} \left( \frac{e^{4x} - 1}{x} - \frac{e^{2x} - 1}{x} \right)$
Apply the limit linearity property.
$\lim_{x \to 0} \frac{e^{4x} - 1}{x} - \lim_{x \to 0} \frac{e^{2x} - 1}{x}$
Apply the standard limit $\lim_{x \to 0} \frac{e^{ax} - 1}{x} = a$ to each term.
$= 4 - 2$ $= 2$
The second limit is:
$\boxed{2}$

Evaluate the following limits without using L'Hôpital's Rule: lim x->0 (6e3x - 6)/(e^x - 1) and lim x->0 (e4x - e2x)/x.