Question

Evaluate limit. $$\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{e^{x}-1}$$

Short answer

Question: Evaluate the limit as `x` approaches 0 for the function \(\frac{e^{4 x}-1}{e^{x}-1}\). Answer: The limit as `x` approaches 0 for the given function is 4.

Step-by-step solution

Make sure the limit is indeterminate form (0/0)

Plug in `x = 0` into the given expression: $$\frac{e^{4 \cdot 0}-1}{e^{0}-1} = \frac{e^{0}-1}{e^{0}-1} = \frac{1-1}{1-1} = \frac{0}{0}$$ The limit is in the indeterminate form \(\frac{0}{0}\).

Apply L'Hopital's Rule - First Derivative

Derive both the numerator and the denominator with respect to `x`: Numerator: \(\frac{d}{dx}(e^{4 x}-1) = 4e^{4 x}\) Denominator: \(\frac{d}{dx}(e^{x}-1) = e^{x}\) Now, we can apply L'Hopital's rule: $$\lim_{x \rightarrow 0} \frac{4e^{4 x}}{e^{x}}$$ Check if we can directly find the limit: $$\lim_{x \rightarrow 0} \frac{4e^{4 \cdot 0}}{e^{0}} = \frac{4e^{0}}{e^{0}} = \frac{4 \cdot 1}{1} = 4$$ The limit can be directly found as the quotient no longer has the indeterminate form. Thus, the solution to the given problem is: $$\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{e^{x}-1} = \boxed{4}$$

Key concepts

Indeterminate Forms

When evaluating limits in calculus, we sometimes encounter what is known as an indeterminate form. These occur when the limit cannot be determined from the initial form of a mathematical expression. A common example is the ratio \( \frac{0}{0} \), which shows up frequently when trying to compute the limits of fractions where both the numerator and the denominator approach zero as the variable approaches a certain value. Other indeterminate forms include \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^{0} \), \( \infty^{0} \), and \( 1^{\infty} \).

To resolve these indeterminate forms, we often employ strategies like factoring, conjugate multiplication, or in particular, L'Hopital's Rule, which uses derivatives to find a limit that eludes direct substitution.

Derivatives

Derivatives play a vital role in calculus, representing the rate of change of a function with respect to a variable. They are fundamental in determining slopes of tangent lines, finding maximum and minimum values of functions, and solving problems involving motion, growth, and approximation. In L'Hopital's Rule, the derivative is used to simplify calculations of limits involving indeterminate forms.

The process of differentiating terms, especially exponential functions like \( e^{nx} \), involves applying the chain rule, which in our case results in \( n \cdot e^{nx} \). For the given problem \( \frac{e^{4x}-1}{e^x-1} \), after identifying that we have an indeterminate form of \( \frac{0}{0} \), we find the derivatives of both the numerator and the denominator with respect to \( x \). This conversion to derivatives allows L'Hopital's Rule to be used effectively to compute the desired limit.

Exponential Functions

Exponential functions, typically in the form of \( e^{x} \) where \( e \) represents Euler's number, are a cornerstone in continuous growth models across various scientific disciplines. One of their remarkable properties is that their rate of change is proportional to the function's value, making the derivative of \( e^{x} \) equal to itself.

In the context of the exercise, exponential growth is observed as \( e^{4x} \) grows much faster than \( e^{x} \) as \( x \) increases. Upon taking derivatives, we notice that the rates at which they grow are distinct but maintain the simplicity that defines exponential functions. It is this simplicity and the exponential function's continuous growth property that enable us to resolve the limit without complication after using L'Hopital's Rule, leading directly to the answer.