Understanding the Concept: Evaluating a Limit

We are given the limit: $\lim_{x \to 0} \frac{e^{2 x} - 1}{x}$ and must find its value. This is a standard limit that evaluates to the derivative of the exponential function at x=0.

Step-by-Step Solution:

Step 1: Recognize the Standard Limit Form

The given limit is of the form $\lim_{h \to 0} \frac{e^{h} - 1}{h} = 1$ . This is a fundamental result.

Step 2: Perform a Substitution

Let $u = 2 x$ . Therefore, as $x \to 0$ , $u \to 0$ as well.

Substituting into our original limit:

$\lim_{x \to 0} \frac{e^{2 x} - 1}{x} = \lim_{u \to 0} \frac{e^{u} - 1}{u / 2}$

Step 3: Simplify the Expression

We can rewrite the denominator:

$\lim_{u \to 0} \frac{e^{u} - 1}{u / 2} = \lim_{u \to 0} 2 \times \frac{e^{u} - 1}{u}$

Step 4: Apply the Standard Limit

We know from the standard result that $\lim_{u \to 0} \frac{e^{u} - 1}{u} = 1$ . Substituting this value:

$2 \times 1 = 2$

Final Answer: The value of the limit is $2$ .

Therefore, the correct option from the images provided is the one that shows the number 2.

Related Topics:

Limits and Continuity
Derivatives (This limit represents the derivative of f(x)=e²ˣ at x=0)
L'Hôpital's Rule (An alternative method to solve this limit)
Exponential Functions

Key Formulae and Theory:

1. Standard Limits:

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

$\lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1$

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

2. L'Hôpital's Rule:

If a limit is in an indeterminate form like 0/0 or ∞/∞, then:

$\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}$

Applying it to our problem: $\frac{d}{d x} (e^{2 x} - 1) = 2 e^{2 x}$ and $\frac{d}{d x} (x) = 1$ . The limit becomes $\lim_{x \to 0} \frac{2 e^{2 x}}{1} = 2 e^{0} = 2$ .

3. Derivative Definition:

This limit is equivalent to f'(0) for the function f(x) = e²ˣ, as $f^{'} (0) = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} = \lim_{h \to 0} \frac{e^{2 h} - 1}{h}$ .