Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{\ln \cos 2 x}{7 x^{2}} $$

Short answer

The limit is \( \frac{-2}{7} \).

Step-by-step solution

Check the Form of the Limit

We begin by substituting \( x = 0 \) into the limit expression \( \frac{\ln \cos 2x}{7x^2} \). This gives us \( \frac{\ln \cos 0}{7 \cdot 0^2} = \frac{\ln 1}{0} = \frac{0}{0} \), which is an indeterminate form.

Differentiate the Numerator and Denominator

Since the limit produces \( \frac{0}{0} \), we can apply l'Hôpital's Rule, which states that if the limit \( \frac{f(x)}{g(x)} \) approaches \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided this second limit exists. Differentiate the numerator: the derivative of \( \ln \cos 2x \) is \( \frac{-2 \sin 2x}{\cos 2x} = -2 \tan 2x \). Differentiate the denominator: the derivative of \( 7x^2 \) is \( 14x \).

Simplify and Substitute Limit Again

The expression under the limit is now \( \lim_{x \to 0} \frac{-2 \tan 2x}{14x} \). Simplify this to \( \lim_{x \to 0} \frac{-\tan 2x}{7x} \). Substitute \( x = 0 \) again to see if the indeterminate form still exists: \( \frac{-\tan 0}{7 \cdot 0} = \frac{0}{0} \), which is still indeterminate.

Apply l'Hôpital's Rule Again

Differentiate the functions a second time: the derivative of \( -\tan 2x \) is \( -2 \sec^2 2x \) and the derivative of \( 7x \) is \( 7 \). Therefore, the limit becomes \( \lim_{x \to 0} \frac{-2 \sec^2 2x}{7} \).

Compute the Final Limit

Evaluate the limit by substituting \( x = 0 \): \( \lim_{x \to 0} \frac{-2 \sec^2 2x}{7} = \frac{-2 \sec^2 0}{7} = \frac{-2 \times 1}{7} = \frac{-2}{7} \). Thus, the limit is \( \frac{-2}{7} \).

Key concepts

Indeterminate Forms

In calculus, not all limits are straightforward to evaluate. Certain conditions can lead to results that are not immediately clear. These conditions are called "indeterminate forms." Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), as well as others like \(0 \times \infty\) and \(1^\infty\). The expression \(\lim_{x \to 0} \frac{\ln \cos 2x}{7x^2}\), initially simplifies to \(\frac{0}{0}\) when we substitute \(x = 0\), indicating an indeterminate form. This is a classic scenario for using mathematical tools such as l'Hôpital's Rule to navigate through the ambiguity and find the true value of the limit.
Determining whether an expression is an indeterminate form is a crucial first step. Before applying l'Hôpital's Rule, always ensure the expression matches one of these indeterminate conditions by substituting the value to which \(x\) approaches.

Differentiation

Differentiation is the process of finding the derivative of a function. It measures how a function changes as its input changes. Derivatives are central to solving limits that result in indeterminate forms, especially when using l'Hôpital's Rule. This rule states that if you encounter a \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) indeterminate form, you can differentiate the numerator and denominator separately to find the limit.
For our function \(\frac{\ln \cos 2x}{7x^2}\), we first find the derivative of the numerator, \(\ln \cos 2x\). Through the chain rule, it is differentiated to \(-2\tan 2x\). The denominator, \(7x^2\), is differentiated to \(14x\). With these new derivatives, we repeatedly apply l'Hôpital's Rule until the expression is no longer indeterminate, facilitating an easier limit evaluation.

• The derivative helps to smooth out functions that are otherwise problematic at specific points.
• Regular practice with derivatives and rules like the chain rule will improve your problem-solving skills with complex limits.

Trigonometric Limits

Trigonometric limits involve functions like sine, cosine, and tangent. These functions can be tricky due to their cyclical nature and the way they behave near certain critical points. In our example of \(\frac{\ln \cos 2x}{7x^2}\), it's important to manipulate trigonometric identities correctly to simplify the limit.
Initially taking the derivative leads to expressions involving \(\tan 2x\), which, evaluated at \(x = 0\), still presents as \(\frac{0}{0}\). A second differentiation transforms our expression to one involving \(\sec^2 2x\). Evaluating this at \(x = 0\) completes the resolution, as \(\sec 0 = 1\), leading to the simple evaluation of a constant term, \(\frac{-2}{7}\).

• Understanding common trigonometric derivatives, such as \(\tan\) and \(\sec\), is valuable for tackling these problems.
• Often, trigonometric expressions need more than one differentiation to resolve indeterminate forms completely.

When dealing with limits involving trigonometric functions, l'Hôpital's Rule helps by methodically reducing these functions to simpler forms, allowing clearer conclusions.