Problem 4 - Calculus Single Variable

Question

Use l'Hôpital's Rule to find the limit, if it exists. \(\lim _{x \rightarrow 0} \frac{x+\sin (5 x)}{x-3 \sin (4 x)}\)

Step-by-Step Solution

Answer

The limit is \(-\frac{6}{11}\).

Step 1: Verify Indeterminate Form

To apply L'Hôpital's Rule, the function must initially be in an indeterminate form of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).Substitute \(x=0\) into the numerator: \[ x + \sin(5x) = 0 + \sin(0) = 0 \]Substitute \(x=0\) into the denominator: \[ x - 3\sin(4x) = 0 - 3\sin(0) = 0 \]Both the numerator and the denominator are 0, so the expression is in the \(\frac{0}{0}\) form.

Step 2: Apply L'Hôpital's Rule

Since the original limit is in the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule, which states that \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \] when the limit on the right exists.Find the derivative of the numerator: \[ \frac{d}{dx} [x + \sin(5x)] = 1 + 5\cos(5x) \]Find the derivative of the denominator: \[ \frac{d}{dx} [x - 3\sin(4x)] = 1 - 12\cos(4x) \]

Step 3: Compute the New Limit

Now, compute the new limit using the derivatives:\[ \lim_{x \to 0} \frac{1 + 5\cos(5x)}{1 - 12\cos(4x)} \]Substitute \(x = 0\):\[ \frac{1 + 5\cos(0)}{1 - 12\cos(0)} = \frac{1 + 5(1)}{1 - 12(1)} = \frac{6}{-11} \]

Step 4: Final Result

The limit exists and after applying L'Hôpital's Rule and evaluating, the limit is:\[ \lim_{x \to 0} \frac{x+\sin (5x)}{x-3\sin (4x)} = \frac{6}{-11} \]

Key Concepts

Understanding Indeterminate Forms

Indeterminate forms are special cases in mathematics that arise when you try to evaluate limits. These forms aren't immediately solvable and include expressions like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). When you substitute a value into both the numerator and denominator of a function, and both result in zero, you have an indeterminate form \( \frac{0}{0} \), which allows for applying special rules like L'Hôpital's Rule.
This rule helps simplify and evaluate these tricky forms by focusing on derivatives rather than the original function itself. Indeterminate forms signify that further analysis is required to determine the limit.
In the context of our exercise, substituting \( x = 0 \) in both the numerator \( x + \sin(5x) \) and the denominator \( x - 3\sin(4x) \) results in zero. Both computations result in an expression of the form \( \frac{0}{0} \), which is why we can proceed to use L'Hôpital’s Rule.

The Role of Calculus Limits

Calculus limits are a fundamental concept in determining the behavior of a function as it approaches a certain point. Limits describe the value that a function approaches as the input approaches some value.
When dealing with expressions, especially rational ones where both the numerator and denominator are functions of a variable approaching zero or infinity, limits help us predict their behavior without actually reaching those points.
In our example, the limit \( \lim_{x \to 0} \frac{x+\sin (5x)}{x-3\sin (4x)} \) requires us to understand what happens to the fraction as \( x \) approaches zero. Using the concept of limits, you explore how close expressions can get to a particular value (often zero or infinity) without being exact. This concept is crucial when tackling expressions involving indeterminate forms.

Applying Derivatives with L'Hôpital's Rule

Derivatives serve as the backbone for L'Hôpital's Rule which is employed to resolve indeterminate forms. The rule says that for functions that result in forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can take the derivative of the numerator and the derivative of the denominator separately, and then find the limit of the resulting function.
In the case of our limit problem, we find the derivative of \( x + \sin(5x) \), which is \( 1 + 5\cos(5x) \), and the derivative of \( x - 3\sin(4x) \), which is \( 1 - 12\cos(4x) \).
Once these derivatives are calculated, we substitute back into the expression and attempt to evaluate the limit. This reduces solving complex problems to arithmetic with limits of the derivatives, making the expression simpler to handle. Hence, after applying the derivatives, we arrive at \( \frac{6}{-11} \) for our original limit approach, delivering a clearer result thanks to the derivative evaluations.