The function f(x) is differentiable at zero and f(0) = 0. Show that for each n such that n is an...

Question:

The function {eq}f(x) {/eq} is differentiable at zero and {eq}f(0) = 0 {/eq}. Show that for each {eq}n {/eq} such that {eq}n {/eq} is an element of {eq}\mathbb{R} {/eq} the following equation is true:

{eq}\lim\limits_{x \rightarrow 0} (\frac{1}{x}(f(x) + f(\frac{x}{2}) + ... + f(\frac{x}{n}))) = (1 +\frac{1}{2} + ... + \frac{1}{n}){f}'(0) {/eq}

Limit:

The limit is a value to which the function approaches as the input approaches some value.

L'hopital's rule is used to find the limit in case of in-determinant forms {eq}\frac{0}{0}\,,\,\frac{\infty }{\infty } {/eq}

A function {eq}f(x) {/eq} is said to be differentiable at {eq}x=a {/eq} if {eq}f'(a) {/eq} exists.

Answer and Explanation:

Write {eq}\lim\limits_{x \rightarrow 0} \left ( \frac{1}{x}\left ( f(x) + f\left ( \frac{x}{2} \right ) + ... + f\left ( \frac{x}{n} \right ) \right )\right ) {/eq} as {eq}\lim\limits_{x \rightarrow 0}\left [ \frac{f(x) + f\left ( \frac{x}{2} \right ) + ... + f\left ( \frac{x}{n} \right )}{x} \right ] {/eq}

Put {eq}x=0 {/eq}

{eq}\lim\limits_{x \rightarrow 0}\left [ \frac{f(x) + f\left ( \frac{x}{2} \right ) + ... + f\left ( \frac{x}{n} \right )}{x} \right ]\\ =\frac{f(0) + f\left ( 0 \right ) + ... + f\left ( 0\right )}{0} \\ =\frac{0}{0} {/eq}

This is an in-determinant form, so, use L'hopital's rule.

Differentiate numerator and denominator of {eq}\left [ \frac{f(x) + f\left ( \frac{x}{2} \right ) + ... + f\left ( \frac{x}{n} \right )}{x} \right ] {/eq}

{eq}\lim\limits_{x \rightarrow 0}\left [ \frac{f(x) + f\left ( \frac{x}{2} \right ) + ... + f\left ( \frac{x}{n} \right )}{x} \right ]\\ =\lim\limits_{x \rightarrow 0}\left [ \frac{f'(x) + \frac{1}{2}f'\left ( \frac{x}{2} \right ) + ... + \frac{1}{n}f'\left ( \frac{x}{n} \right )}{1} \right ]\\ =\lim\limits_{x \rightarrow 0}f'(x) + \frac{1}{2}f'\left ( \frac{x}{2} \right ) + ... + \frac{1}{n}f'\left ( \frac{x}{n} \right )\\ =f'(0) + \frac{1}{2}f'\left ( 0\right ) + ... + \frac{1}{n}f'\left ( 0 \right ) {/eq}

As {eq}f {/eq} is differentiable at {eq}x=0 {/eq}, {eq}f'(0) {/eq} exists.

{eq}\lim\limits_{x \rightarrow 0} \left ( \frac{1}{x}\left ( f(x) + f\left ( \frac{x}{2} \right ) + ... + f\left ( \frac{x}{n} \right ) \right )\right )\\ =f'(0) + \frac{1}{2}f'\left ( 0\right ) + ... + \frac{1}{n}f'\left ( 0 \right )\\ =f'(0)\left [1+\frac{1}{2}+...+ \frac{1}{n} \right ] {/eq}


Learn more about this topic:

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When to Use the Quotient Rule for Differentiation

from Math 104: Calculus

Chapter 8 / Lesson 8
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