×

# Please I need help in that limit proof?

Please can anybody help me in proving this limit identity? $\large \lim_{x \to a}{\frac{x^n - a^n}{x^m - a^m}} = \frac{n}{m}(a)^{n - m}$

Note by Mohamed Ahmed Abd El-Fattah
2 years, 7 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Note $$x^k-a^k$$ factors as $$\displaystyle(x-a)\sum_{i=1}^kx^{k-i}a^{i-1}$$.

Then $$\displaystyle\lim_{x\to a}\frac{x^n-a^n}{x^m-a^m}=\lim_{x\to a}\frac{(x-a)\sum_{i=1}^nx^{n-i}a^{i-1}}{(x-a)\sum_{i=1}^mx^{m-i}a^{i-1}}$$

$$\displaystyle=\lim_{x\to a}\frac{\sum_{i=1}^nx^{n-i}a^{i-1}}{\sum_{i=1}^mx^{m-i}a^{i-1}}=\frac{\sum_{i=1}^na^{n-i}a^{i-1}}{\sum_{i=1}^ma^{m-i}a^{i-1}}$$

$$\displaystyle=\frac{na^{n-1}}{ma^{m-1}}=\frac{n}{m}a^{n-m}$$.

- 2 years, 7 months ago

Calculus approach:

$\dfrac{x^n - a^n}{x^m - a^m} = \dfrac{x^n - a^n}{x-a} \cdot \dfrac{x-a}{x^m - a^m} = \dfrac{x^n - a^n}{x-a} \div \dfrac{ x^m - a^m}{x-a}$

Applying the limit:

$\begin{eqnarray} \lim_{x\to a} \dfrac{x^n - a^n}{x^m - a^m} &=& \lim_{x\to a} \dfrac{x^n - a^n}{x-a} \div \dfrac{ x^m - a^m}{x-a} \\ &=& \left ( \left . \frac d{dx} (x^n) \right |_{x=a} \right ) \div \left( \left . \frac d{dx} (x^m) \right |_{x=a} \right ) \\ &=& \left [ n a^{n-1} \right ] \div \left [m a^{m-1} \right ] \\ &=& \dfrac nm a^{n-m} \\ \end{eqnarray}$

- 2 years, 7 months ago

Oh right, you could just apply L'Hopital directly:

$$\displaystyle \lim_{x\to a}\frac{x^n-a^n}{x^m-a^m}=\lim_{x\to a}\frac{\frac{d}{dx}(x^n-a^n)}{\frac{d}{dx}(x^m-a^m)}=\lim_{x\to a}\frac{nx^{n-1}}{mx^{m-1}}$$

$$\displaystyle =\lim_{x\to a}\frac{n}{m}x^{n-m}=\frac{n}{m}a^{n-m}$$.

- 2 years, 7 months ago

I got it ^_^ . Thank you very much for your useful help & sharing thoughts :) .

- 2 years, 7 months ago

Thank you very much . Really, very nice approach ^_^ .

- 2 years, 7 months ago