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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, 3 months ago

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

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

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

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

Maggie Miller - 2 years, 3 months ago

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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:

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

Pi Han Goh - 2 years, 3 months ago

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

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

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

Maggie Miller - 2 years, 3 months ago

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

Mohamed Ahmed Abd El-Fattah - 2 years, 3 months ago

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

Mohamed Ahmed Abd El-Fattah - 2 years, 3 months ago

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