Problem 59 - Calculus Single Variable

Question

Relate the limit $$ \lim _{h \rightarrow 0} \frac{e^{h / 2}-e^{-h / 2}}{h} $$ to a derivative and evaluate it.

Step-by-Step Solution

Answer

The limit evaluates to 2, related to the derivative of \(\sinh(x)\) at zero.

Step 1: Identify the Expression as a Derivative

The limit provided can be associated with the definition of a derivative. Recall that the derivative of a function at a point can be expressed as a limit: \(f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\). Compare the given limit to this form to identify the function \(f(x)\) and the point. Notice that the expression \( e^{h/2} - e^{-h/2} \) indicates symmetry around zero for the function \( f(x) \).

Step 2: Simplify the Expression

Observe that \( e^{h/2} - e^{-h/2} \) is the difference of exponentials, suggesting it could be twice the hyperbolic sine function: \( 2\sinh(x) = e^{x} - e^{-x} \). Therefore, the expression given becomes: \( 2\sinh(h/2) \). Thus, the limit can be rewritten as \( \lim_{h \rightarrow 0} \frac{2\sinh(h/2)}{h} \).

Step 3: Relate to the Definition of the Derivative

Consider \( f(x) = \sinh(x) \) so that the derivative \( f'(x) = \cosh(x) \). Since \( f(x) = \sinh(x) \), we find that the limit \( \lim_{h \rightarrow 0} \frac{\sinh(h/2)}{h/2} = 1 \) because \( \frac{\sinh(u)}{u} \rightarrow 1 \) as \( u \rightarrow 0 \). This involves substituting \( u = h/2 \), hence \( \lim_{h \rightarrow 0} \frac{\sinh(h/2)}{h/2} = 1 \).

Step 4: Calculate the Final Value

Returning to our expression \( \lim_{h \rightarrow 0} \frac{2\sinh(h/2)}{h} \), rewrite it as \( 2 \times \frac{\sinh(h/2)}{h/2} \). Given that this portion tends to 1, the entire expression evaluates to \( 2 \times 1 = 2 \).

Step 5: State the Relation to the Derivative

Thus, the expression can be seen as evaluating the limit \( \lim_{h \rightarrow 0} \frac{f(h) - f(0)}{h} \) of a function \( f(x) = \sinh(x) \), whose derivative at zero results in the hyperbolic cosine function \( \cosh(x) \) evaluated at zero, which is \(\cosh(0) = 1\) when adjusted for scaling factors.

Key Concepts

Limit of a Function

The concept of the limit is fundamental in calculus. It helps us determine the value that a function approaches as the input gets closer to a certain point. In the given exercise, we deal with a special kind of limit which relates to the derivative of a function. To understand this, it's crucial to recall the definition of a derivative. The derivative of a function at a particular point is defined as the limit: \[f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\]This formula represents the rate of change of the function near that point. In the exercise, our objective is to relate the given limit to this definition. To do this, we compare the given limit expression with the derivative form. By identifying the function part in our limit to match it with \( f(x) \), we can evaluate the derivative effectively. In essence, when evaluating limits that resemble derivatives, remember:

• Replace the difference expression with a difference quotient form, \( f(x+h) - f(x) \).
• Determine \( h \rightarrow 0 \) to find the instantaneous rate of change.
• Recognize patterns in expressions involving standard functions such as exponential and hyperbolic functions.

With these steps, you can break down complex limits into manageable forms and understand their implications in terms of derivatives.

Hyperbolic Functions

Hyperbolic functions include common functions like hyperbolic sine (\(\sinh \)) and hyperbolic cosine (\(\cosh \)). They are analogous to the trigonometric functions but are based on hyperbolas rather than circles.The hyperbolic sine function is defined as:\[\sinh(x) = \frac{e^x - e^{-x}}{2}\]Similarly, the hyperbolic cosine function is:\[\cosh(x) = \frac{e^x + e^{-x}}{2}\]These functions have unique properties:

• The derivative of \(\sinh(x)\) is \(\cosh(x)\), paralleling the relationship between \(\sin(x)\) and \(\cos(x)\).
• They model hyperbolic shapes and appear in various areas of mathematics and physics.

In the exercise, we identify \( e^{h/2} - e^{-h/2} = 2\sinh(h/2) \), noticing how the expression fits perfectly into the form of the hyperbolic sine function.
By understanding hyperbolic functions, you gain insights into how these expressions can simplify seemingly complex limit problems. Recognizing these patterns is crucial in evaluating limits and derivatives tied to hyperbolic functions.

Exponential Functions

Exponential functions are functions in which the variable appears as an exponent. A key example is the natural exponential function \( e^x \), where \( e \) is the base of the natural logarithms, approximately equal to 2.718.These functions grow at rates proportional to their current value, leading to rapid increases or decreases. Exponential functions have important properties:

• The derivative of \(e^x\) is \(e^x\), showing the function's self-similar nature.
• They model continuous growth, decay, and appear in various phenomena like population growth and radioactive decay.

In the given exercise, the components \( e^{h/2} \) and \( e^{-h/2} \) suggest an exponential pattern. Such forms can lead to hyperbolic functions when expressed in combinations like \( e^x - e^{-x} \). Simplifying these expressions requires an understanding of exponential properties.
By mastering exponential functions, you can unlock strategies to tackle many mathematically rich real-world problems, harnessing their power to derive other essential functions like hyperbolic ones.