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Find the limit, if it exists. $$ \lim _{x \rightarrow 0} \frac{2-e^{x}-e^{-x}}{1-\cos ^{2} x} $$

Short Answer

Expert verified
The limit is -1.

Step by step solution

01

Identify the Indeterminate Form

First, substitute \( x = 0 \) into the expression to check if it forms an indeterminate form. The numerator becomes \( 2 - e^0 - e^0 = 2 - 1 - 1 = 0 \), and the denominator becomes \( 1 - \cos^2(0) = 0 \). So, the expression \( \frac{0}{0} \) is an indeterminate form.
02

Simplify the Denominator

Recognize that \( \cos^2(x) = 1 - \sin^2(x) \), so the denominator \( 1 - \cos^2(x) \) simplifies to \( \sin^2(x) \). The limit becomes \( \lim_{x \to 0} \frac{2 - e^x - e^{-x}}{\sin^2(x)} \).
03

Apply L'Hôpital's Rule

Since we have the indeterminate form \( \frac{0}{0} \), apply L'Hôpital's Rule, which involves differentiating the numerator and denominator separately. The derivative of the numerator \( 2 - e^x - e^{-x} \) is \( -e^x + e^{-x} \). The derivative of the denominator \( \sin^2(x) \) is \( 2\sin(x)\cos(x) \) or \( \sin(2x) \).
04

Compute the New Limit

The new limit is \( \lim_{x \to 0} \frac{-e^x + e^{-x}}{\sin(2x)} \). Substitute \( x = 0 \) to check: the numerator \( -e^0 + e^0 = 0 \), and \( \sin(2 \times 0) = 0 \) in the denominator leads to \( \frac{0}{0} \) again. Thus, we apply L'Hôpital's Rule once more.
05

Apply L'Hôpital's Rule Again

Differentiate the numerator and denominator again. The derivative of the numerator \( -e^x + e^{-x} \) is \( -e^x - e^{-x} \). The derivative of \( \sin(2x) \) is \( 2\cos(2x) \). The limit now is \( \lim_{x \to 0} \frac{-e^x - e^{-x}}{2\cos(2x)} \).
06

Evaluate the Final Limit

Substitute \( x = 0 \) into the new expression: the numerator is \( -e^0 - e^0 = -2 \), and the denominator is \( 2 \cos(0) = 2 \). Thus, the limit is \( \frac{-2}{2} = -1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Indeterminate Forms
In calculus, an indeterminate form often arises when evaluating limits. These forms occur when the direct substitution of a value into a limit results in an ambiguous expression, such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 0^0 \).
These forms don't immediately provide a clear indication of the behavior of a function as it approaches a point, making the limit difficult to evaluate without further investigation.

In the provided exercise, the initial substitution resulted in \( \frac{0}{0} \), which is a classic example of an indeterminate form.
Since both the numerator and the denominator approach zero as \( x \) approaches zero, more analytical techniques, such as L'Hôpital's Rule or algebraic manipulation, are required to resolve the limit.
  • The indeterminate form invites further differentiation or algebraic simplification.
  • Careful handling is crucial to avoid drawing incorrect conclusions from the initial substitution alone.
Grasping Limits in Calculus
Limits in calculus are foundational, serving as the stepping stone to more complex concepts like continuity and derivatives. They help us understand the behavior of functions as they approach a specific point.
Instead of determining the value of a function at a point, limits tell us what value a function approaches.

The expression \( \lim _{x \to 0} \frac{2 - e^x - e^{-x}}{1 - \cos^2{x}} \) examined in this exercise uses limits to understand the behavior of the fractional expression as \( x \) approaches zero.
  • Limits can reveal much about a function's nature, even if the function isn't defined at that point.
  • They can be computed using various techniques, such as substitution, graphical analysis, or algebraic simplification.
  • When limits result in indeterminate forms, additional methods like L'Hôpital's Rule may be needed to evaluate them.
The Role of Differentiation
Differentiation is a fundamental tool in calculus, used to compute the derivative of a function. Derivatives represent the rate at which a quantity changes, providing insights into the behavior and nature of functions.
In the context of limits, particularly indeterminate forms, differentiation can be applied through L'Hôpital's Rule to evaluate limits that initially seem undefined.

For instance, in the original solution, differentiation was applied to both the numerator and the denominator of the fraction to resolve the indeterminate form \( \frac{0}{0} \). L'Hôpital's Rule states that:
  • If \( \lim _{x \to a} f(x) = \lim _{x \to a} g(x) = 0 \) or \( \pm \infty \), then:

    \( \lim _{x \to a} \frac{f(x)}{g(x)} = \lim _{x \to a} \frac{f'(x)}{g'(x)} \) provided the right-side limit exists.
  • This method is extremely useful when faced with indeterminate forms as it often clarifies the limit.
By applying differentiation correctly, one can navigate through complex limits to find deterministic results, like \(-1\) in this exercise.

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