Understanding the Concept: Evaluating a Limit

The given problem is to evaluate the limit: $\lim_{x \to 0} \frac{e^{3 x} - e^{2 x}}{x}$ . This is a classic $\frac{0}{0}$ indeterminate form, which we can solve using the standard limit result for the exponential function.

Step-by-Step Solution

Step 1: Identify the Indeterminate Form
Substitute $x = 0$ into the expression: $\frac{e^{3 \times 0} - e^{2 \times 0}}{0} = \frac{e^{0} - e^{0}}{0} = \frac{1 - 1}{0} = \frac{0}{0}$ This confirms it is an indeterminate form, so we must apply a limit rule.

Step 2: Recall the Standard Limit
The fundamental limit we use is: $\lim_{t \to 0} \frac{e^{t} - 1}{t} = 1$ Our goal is to manipulate our given limit to match this form.

Step 3: Manipulate the Given Expression
We can split the numerator and rewrite the expression: $\frac{e^{3 x} - e^{2 x}}{x} = \frac{e^{2 x} (e^{x} - 1)}{x} = e^{2 x} \times \frac{e^{x} - 1}{x}$

Step 4: Apply the Limit
Now, we take the limit of the entire expression as $x \to 0$ : $\lim_{x \to 0} (e^{2 x} \times \frac{e^{x} - 1}{x})$ Since this is a product of two functions, we can apply the limit: $(\lim_{x \to 0} e^{2 x}) \times (\lim_{x \to 0} \frac{e^{x} - 1}{x})$

Step 5: Evaluate Each Limit
Let's solve each part: $\lim_{x \to 0} e^{2 x} = e^{2 \times 0} = e^{0} = 1$ $\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$ (This is the direct application of the standard limit).

Final Answer:
Therefore, the value of the limit is: $\lim_{x \to 0} \frac{e^{3 x} - e^{2 x}}{x} = 1 \times 1 = 1$

Related Topics

Limits and Continuity
Indeterminate Forms ( $\frac{0}{0}$ , $\frac{\infty}{\infty}$ )
L'Hôpital's Rule (An alternative method to solve such limits)
Derivatives (The standard limit $\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$ is actually the derivative of $e^{x}$ at x=0).

Key Formulae & Theory

Standard Limits:

$\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$
$\lim_{x \to 0} \frac{\sin x}{x} = 1$
$\lim_{x \to 0} \frac{\log (1 + x)}{x} = 1$

Properties of Limits:

The limit of a product is the product of the limits: $\lim_{x \to a} (f (x) \cdot g (x)) = (\lim_{x \to a} f (x)) \cdot (\lim_{x \to a} g (x))$ , provided the individual limits exist.