Question

Evaluate the following limits or state that they do not exist. $$\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin ^{2} x}$$

Short answer

Answer: No, the limit does not exist.

Step-by-step solution

Identify the type of limit

We are given the limit: $$\lim _{x \rightarrow 0} \frac{\cos x-1}{\sin ^{2} x}$$ Our task is to evaluate this limit as x approaches 0.

Expand and simplify the expression

Let's first rewrite the expression using the double angle identity for cosine, which is: $$\cos(2x)=2\cos^2(x)-1$$ Now, we can rewrite our expression as: $$\lim _{x \rightarrow 0} \frac{-\frac{1}{2} + \frac{1}{2} \cos(2x)}{\sin^2(x)}$$

Apply the double-angle identity for sine

Next, we change the denominator using the double-angle identity for sine: $$\sin(2x) = 2 \sin(x) \cos(x)$$ Since \(\sin^2(x) = (\sin(x))^2\), we can write it as: $$\sin^2(x) = (\frac{1}{2} \sin(2x))^2 = \frac{1}{4} \sin^2(2x)$$ Now, our expression is: $$\lim _{x \rightarrow 0} \frac{-\frac{1}{2} + \frac{1}{2} \cos(2x)}{\frac{1}{4}\sin^2(2x)}$$

Simplify and apply the limit

Simplify the expression by multiplying the numerator and denominator by 4 (to eliminate the fraction): $$\lim _{x \rightarrow 0} \frac{4\left(-\frac{1}{2} + \frac{1}{2} \cos(2x)\right)}{\sin^2(2x)}$$ Simplify further: $$\lim _{x \rightarrow 0} \frac{-2 + 2\cos(2x)}{\sin^2(2x)}$$ Now, we apply the limit to the expression, using the fact that \(\lim_{x \rightarrow 0}\cos(x) = 1\) and \(\lim_{x \rightarrow 0}\sin(x) = 0\): $$\lim _{x \rightarrow 0} \frac{-2 + 2\cos(2(0))}{\sin^2(2(0))}$$ After calculating: $$\lim _{x \rightarrow 0} \frac{-2 + 2\cos(0)}{\sin^2(0)}$$ We have: $$\lim _{x \rightarrow 0} \frac{-2 + 2 \cdot 1}{0}$$ As we can see, the denominator is zero, which means the limit does not exist.

Key concepts

Trigonometric limits

Trigonometric limits often involve functions like sine, cosine, or tangent, and can be particularly interesting because they require a good understanding of trigonometric identities and limit properties. These limits become challenging when the trigonometric function results in expressions that involve 0/0, which is an indeterminate form. To approach these problems, we often use trigonometric identities to simplify the expression. Key identities include:

• The Pythagorean identity: \(\sin^2(x) + \cos^2(x) = 1\)
• Double angle identities, like \(\sin(2x) = 2\sin(x)\cos(x)\) and \(\cos(2x) = 2\cos^2(x) - 1\)
• Half-angle and sum-to-product identities

By applying these identities, we can transform the trigonometric expression into a form that is easier to evaluate when taking the limit, helping us bypass the indeterminate form.

Limit evaluation

Evaluating limits involves determining the value that a function approaches as the input approaches a particular point. The exercise provided deals with a limit evaluation where the function involves trigonometric terms. The main technique involves utilizing mathematical identities and simplifications to handle complicated expressions. When evaluating a limit, especially when 0 over 0 is present, you should:

• Identify the form of the expression at the limit point (often, it may not be immediately solvable due to indeterminate forms)
• Simplify the expression using algebraic and trigonometric identities
• Substitute to see if the limit can be directly evaluated after simplification

In our exercise, we have used these steps. The original expression was simplified using a series of identities, attempting to make further mathematical manipulations possible so we can interpret the physical behavior of the function near the approaching value. Although simplifications helped clear some parts of the problem, it ultimately led to a conclusion that the limit does not exist because the mathematics suggested a division by zero—an undefined operation.

Indeterminate forms

Indeterminate forms are expressions in limits that do not initially give us clear information about what value the limit might approach. Typical examples include \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). Such forms mean straightforward substitution doesn't directly yield a usable result.In cases of indeterminate forms, several strategies are used to clarify the value of the limit, such as:

• L'Hôpital's rule: Applicable when the limit results in a 0/0 or ∞/∞ form
• Simplification through algebraic manipulation or trigonometric identities
• Expansion using Taylor or Maclaurin series for approximations

The provided exercise reached an indeterminate form, where the direct evaluation wasn't possible because both numerator and denominator approached zero. Simplification transformed it somewhat but ultimately ended at a state where the limit didn’t exist, highlighting the importance of identifying and appropriately handling indeterminate forms.