Problem 63 - Single Variable Calculus

Question

Show that if \(f^{\prime}(x)\) exists, then \(\lim _{h \rightarrow 0} \frac{f(x+n h)-f[x+(n-1) h]}{h}=f^{\prime}(x) \quad n \neq 0,1\)

Step-by-Step Solution

Answer

We are given the expression \(\lim _{h \rightarrow 0} \frac{f(x+n h)-f[x+(n-1) h]}{h}\) and asked to show that it is equal to \(f'(x)\) if the derivative exists and \(n \neq 0, 1\). To do this, we substitute \(t = nh\), apply the limit properties, and rewrite the expression in terms of the definition of the derivative. After canceling the n term from both sides, we obtain \(f'(x) = \lim_{t \rightarrow 0} \frac{f(x+t)-f(x+t-nh)}{t}\), proving that the given limit expression is equal to \(f'(x)\) when the derivative exists and \(n \neq 0, 1\).

Step 1: Write down the definition of derivative f'(x)

Recall that the definition of the derivative of a function f at the point x is given by, \[ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \]

Step 2: Manipulate the provided expression

The given expression is, \[ \lim _{h \rightarrow 0} \frac{f(x+n h)-f[x+(n-1) h]}{h} \] To prove that this is equal to f'(x), we will rewrite the expression in terms of the definition of the derivative.

Step 3: Substitute t = nh in the expression

Let's introduce a substitution to simplify the expression. Let t = nh, so that when h approaches 0, t also approaches 0. Now, in terms of t, the expression becomes, \[ \lim_{t \rightarrow 0} \frac{f(x+t)-f(x+t-nh)}{t} \cdot \frac{n}{n} \]

Step 4: Apply the limit properties

Since both n and t are approaching 0, we can apply the limit properties as follows, \[ \lim_{t \rightarrow 0} n \cdot \frac{f(x+t)-f(x+t-nh)}{t} = n \cdot \lim_{t \rightarrow 0} \frac{f(x+t)-f(x+t-nh)}{t} \]

Step 5: Rewrite the expression in terms of f'(x)

Now, rewrite the expression in terms of f'(x) using the definition of the derivative and the fact that t approaches 0 as h approaches 0, \[ n \cdot f'(x) = n \cdot \lim_{t \rightarrow 0} \frac{f(x+t)-f(x+t-nh)}{t} \]

Step 6: Cancel the n term from both sides

Since n is not equal to 0 or 1, we can cancel the n term from both sides of the equation. This gives, \[ f'(x) = \lim_{t \rightarrow 0} \frac{f(x+t)-f(x+t-nh)}{t} \] Thus, we have proved that if the derivative f'(x) exists, the given limit expression is equal to f'(x) when n is not equal to 0 or 1.

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus. It measures how a function changes as its input changes. In simpler terms, it tells us the rate of change or the slope of a function at any given point.

Think of it like this: if you have a car moving along a road, the speedometer tells you the instantaneous speed, which is analogous to the derivative. For a function \( f(x) \), the derivative at a point \( x \), written as \( f'(x) \), is defined by the limit:

• \( f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \)

This definition means that we're looking at how the function value changes as we make tiny changes, even infinitesimal, to \( x \).

If this limit exists, it tells us exactly how steep the function is at \( x \). This is crucial because it helps in understanding the behavior of functions and solving many real-world problems.

Limit

A limit in calculus helps us understand the behavior of a function as it approaches a particular point. It's a way of expressing the value that a function approaches as the input gets closer to some number.

Mathematically, to determine the limit of \( f(x) \) as \( x \) approaches a certain point \( a \), we use the notation:

• \( \lim_{x \to a} f(x) \)

This tells us what \( f(x) \) gets closer to as \( x \) moves closer to \( a \), without ever necessarily touching \( a \).

In the exercise, we work with the limit as \( h \) goes to zero, which is key to finding a derivative. Understanding limits is pivotal because they form the groundwork for defining derivatives and integrals.

Substitution

Substitution is a technique often used in calculus to simplify complex expressions or integrals. It involves replacing a part of one expression with another to make it easier to manipulate.

In the exercise solution, substitution was applied by letting \( t = nh \). This was strategically done to transform the expression into a form that resembles the derivative.

• Why substitution? It often rescales parameters or redefines them in simpler terms.
• How it's used? You replace the variables in an expression with new ones and adjust the limits accordingly.

Using substitution smartly can provide clarity and simplicity, enabling us to solve limits and derivatives more effectively.

Limit Properties

Limit properties are rules that allow us to manipulate and simplify limits. These are essential tools in calculus when evaluating complex limit expressions.

Some common properties include:

• Scalar multiple: \( \lim_{x \to a} c \cdot f(x) = c \cdot \lim_{x \to a} f(x) \)
• Sum/Difference: \( \lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \)
• Product: \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \)

In the exercise, the property of distributing a scalar over a limit was employed to extract \( n \) from the limit expression, which helped simplify it down to a derivative form.

By understanding and applying these properties, we can tackle even complicated limit calculations with greater ease.