Problem 55 - Calculus Single Variable

Question

Calculate the given limit. \(\lim _{x \rightarrow 0} \frac{\sinh (x)}{\exp (x)-1}\)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand Hyperbolic Sine Function

Recall the definition of the hyperbolic sine function: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).

Step 2: Express the Numerator

Substitute the definition of \( \sinh(x) \) into the limit expression: \[ rac{\sinh(x)}{\exp(x)-1} = \frac{\frac{e^x - e^{-x}}{2}}{e^x - 1} \].

Step 3: Simplify the Fraction

Cancel out \(\frac{1}{2}\) in the numerator and rewrite the expression:\[ \frac{e^x - e^{-x}}{2(e^x - 1)} \].

Step 4: Apply L'Hôpital's Rule

The limit is of the indeterminate form \(\frac{0}{0}\) as \(x \to 0\), so apply L'Hôpital's Rule by differentiating the numerator and the denominator:Numerator derivative: \((e^x + e^{-x}) = e^x - (-e^{-x}) \),Denominator derivative: \( e^x \).

Step 5: Retry the Limit

Substitute back into the limit to evaluate:\[ \lim_{x \to 0} \frac{e^x + e^{-x}}{2e^x} \].

Step 6: Simplify Further

Simplify the limit expression:\[ = \frac{1}{2} \lim_{x \to 0} \left(1 + \frac{e^{-x}}{e^x} \right) \].Since \(e^{-x}/e^x = e^{-2x}\), as \(x \to 0, e^{-2x} \to 1\).

Step 7: Evaluate the Final Limit

Evaluate the simplified limit:\[ = \frac{1}{2}(1 + 1) = \frac{1}{2}(2) = 1 \].

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are defined using exponential functions. The hyperbolic sine function, denoted as \(\sinh(x)\), is defined as:\[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]This function is closely related to exponential growth and decay, making it useful in various fields such as physics and engineering. Unlike trigonometric functions, which are periodic, hyperbolic functions deal with hyperbolas and exhibit similar properties such as symmetry.

• \(\sinh(x)\) is an odd function: \(\sinh(-x) = -\sinh(x)\)
• The inverse function is \(\operatorname{arsinh}(x)\) or \(\sinh^{-1}(x)\).

Understanding these functions is essential for solving problems involving limits and derivatives, as they appear frequently in calculus involving exponential expressions.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for calculating limits, especially when facing indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The rule states that:If \(\lim_{x \to c} \frac{f(x)}{g(x)}\) results in an indeterminate form, then:\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]provided the derivatives \(f'(x)\) and \(g'(x)\) exist and \(g'(x) eq 0\) around the point \(c\), except possibly at \(c\).In our exercise, we applied L'Hôpital's Rule because the limit yielded \(\frac{0}{0}\) when plugging in \(x = 0\). Upon differentiating both the numerator and the denominator, the new expression allows us to evaluate the limit directly, often simplifying the problem significantly.
L'Hôpital's Rule is especially useful when direct substitution leads to complex expressions that are difficult to simplify by other means.

Indeterminate Forms

In calculus, indeterminate forms arise when evaluating limits that initially seem undefined. Common examples include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and forms like \(0 \times \infty\). These expressions hint at a need for further analysis to find an actual value.

• \(\frac{0}{0}\) suggests overlapping changes in both numerator and denominator, requiring differentiation or factorization.
• \(\frac{\infty}{\infty}\) might simplify by using L'Hôpital's Rule to tackle asymptotic behavior.

In the given problem, we encountered an indeterminate form of \(\frac{0}{0}\) when substituting \(x = 0\). Recognizing this allowed us to apply L'Hôpital’s Rule to effectively determine the limit. Indeterminate forms are fundamental in calculus for addressing situations where straightforward approaches like substitution do not work, revealing more about the behavior of functions at specific points.