Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{\tan x-x}{\sin 2 x-2 x} $$

Short answer

The limit is \( -\frac{1}{4} \).

Step-by-step solution

Check for Indeterminate Form

First, substitute \( x = 0 \) into the given expression to check if it forms an indeterminate type. Substitute the values in the limit expression: \( \frac{\tan x - x}{\sin 2x - 2x} \). Both \( \tan 0 - 0 = 0 \) and \( \sin(2 \times 0) - 2 \times 0 = 0 \), which forms \( \frac{0}{0} \), an indeterminate form.

Apply l'Hôpital's Rule

Since the expression is in the \(\frac{0}{0}\) indeterminate form, we can apply l'Hôpital's Rule, which tells us to take the derivative of the numerator and denominator separately. Differentiate the numerator: \( \frac{d}{dx}(\tan x - x) = \sec^2 x - 1 \). Differentiating the denominator \( \frac{d}{dx}(\sin 2x - 2x) = 2\cos 2x - 2 \).

Reevaluate the Limit

Reformulate the limit using the derivatives obtained: \( \lim_{x \rightarrow 0} \frac{\sec^2 x - 1}{2\cos 2x - 2} \). Substitute \( x = 0 \) in the derivative-filled expression: \( \frac{\sec^2(0) - 1}{2\cos(0) - 2} \). This simplifies to \( \frac{1^2 - 1}{2 \times 1 - 2} = \frac{0}{0} \). It is still an indeterminate form, so we apply l'Hôpital's Rule again.

Apply l'Hôpital's Rule Again

Taking another derivative of the numerator and denominator: The derivative of \( \sec^2 x - 1 \) is \( 2\sec^2 x \tan x \); the derivative of \( 2\cos 2x - 2 \) is \( -4\sin 2x \).

Evaluate the Limit After Second Derivative

Now, substitute \( x = 0 \) again: \( \lim_{x \rightarrow 0} \frac{2\sec^2 x \tan x}{-4\sin 2x} \). Substitute \( x = 0 \) in this expression: \( \frac{2(1) \times 0}{-4 \times 0} = \frac{0}{0} \) is encountered again. To fully resolve, recognize that derivatives are constant around zero. Consider third-order terms of Taylor series for an exact approach or simplify by recognizing previous forms not possible.

Simplify Using Taylor Series Expansion

Expand \( \tan x \) and \( \sin 2x \) using Taylor series around \( x = 0 \): \( \tan x \approx x + \frac{x^3}{3} + O(x^5) \) and \( \sin 2x \approx 2x - \frac{(2x)^3}{6} + O(x^5) \). Substitute back into the original expression, \( \frac{\tan x - x}{\sin 2x - 2x} \approx \frac{\frac{x^3}{3}}{-\frac{8x^3}{6}} = \frac{x^3 / 3}{-4x^3 / 3} \approx \frac{1}{-4} = -\frac{1}{4}.\)

Conclusion: Find the Result

The limit is therefore \( -\frac{1}{4} \). After applying l'Hôpital's Rule twice and using Taylor series for further simplification, the limit is finally resolved.

Key concepts

l'Hôpital's Rule

When faced with a limit problem that exists in an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), l'Hôpital's Rule becomes highly useful. This rule helps by allowing us to differentiate the numerator and the denominator of the problematic fraction separately. The condition for using l'Hôpital's Rule is quite clear:

• Your original limit must be in an indeterminate form.
• Both numerator and denominator must be differentiable near the point of interest.

Applying the rule transforms the limit into a simpler form by applying derivatives, thereby making it easier to evaluate. For our exercise, l'Hôpital's Rule was applied twice, which is common when the limit remains indeterminate after the first application.

Indeterminate Forms

In calculus, indeterminate forms are expressions that initially appear undefined or ambiguous. A classic example is \( \frac{0}{0} \) or forms involving \( \infty \) such as \( \frac{\infty}{\infty} \). When you plug numbers into a limit expression and end up with these forms, it indicates more work needs to be done to settle on a concrete value, often through a mathematical tool or simplification.Why do these forms occur? Often, the behavior of the function near the point of evaluation is more complex, hinting at subtle cancellations or relationships that aren't immediately obvious. In our exercise, before applying l'Hôpital's Rule, recognizing this form is crucial. It assures the problem is set up correctly for further techniques like differentiation or series expansion.

Taylor Series

The Taylor series is a powerful tool in approximating functions with polynomials that match the function's values and derivatives at a single point, typically zero (also known as the Maclaurin series).Using a Taylor series in limit calculations allows for simplification when functions appear intractable at first. In our problem, the Taylor expansion of \( \tan x \) and \( \sin 2x \) near zero offered a straightforward approximation:

• \( \tan x \approx x + \frac{x^3}{3} + O(x^5) \)
• \( \sin 2x \approx 2x - \frac{8x^3}{6} + O(x^5) \)

By substituting these into the expression and simplifying, we achieve a clear, non-zero limit result. This step sometimes helps avoid repeated use of l'Hôpital's Rule when higher precision or simplification is necessary.

Derivatives in Limits

Derivatives play a crucial role in computing limits, especially when applying l'Hôpital's Rule. Finding derivatives refines a function's form, leading to a numerically stable expression around a point of interest, say zero in our case.Differentiating can handle tricky expressions by reducing the complexity. For instance, in the exercise, differentiating the numerator \( \tan x - x \) and denominator \( \sin 2x - 2x \) helped shed the problematic \( 0/0 \) form. In iterative l'Hôpital's applications:

• The first derivative turns \( \frac{\tan x - x}{\sin 2x - 2x} \) into simpler terms \( \frac{\sec^2 x - 1}{2\cos 2x - 2} \).
• Further differentiation leads to \( \frac{2\sec^2 x \tan x}{-4\sin 2x} \), still requiring attention.

Thus, derivatives remain key when refining limits, matching Taylor series expansions, and resolving studied mathematical expressions.