Please I need help in that limit proof?

Please can anybody help me in proving this limit identity? lim⁡x→axn−anxm−am=nm(a)n−m \large \lim_{x \to a}{\frac{x^n - a^n}{x^m - a^m}} = \frac{n}{m}(a)^{n - m}

#Calculus #Limits

Note by Mohamed Ahmed Abd El-Fattah
5 years, 10 months ago

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1 vote

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Comments

Note xk−akx^k-a^k factors as (x−a)∑i=1kxk−iai−1\displaystyle(x-a)\sum_{i=1}^kx^{k-i}a^{i-1}.

Then lim⁡x→axn−anxm−am=lim⁡x→a(x−a)∑i=1nxn−iai−1(x−a)∑i=1mxm−iai−1\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}}

=lim⁡x→a∑i=1nxn−iai−1∑i=1mxm−iai−1=∑i=1nan−iai−1∑i=1mam−iai−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}}

=nan−1mam−1=nman−m\displaystyle=\frac{na^{n-1}}{ma^{m-1}}=\frac{n}{m}a^{n-m}.

Maggie Miller - 5 years, 10 months ago

Calculus approach:

xn−anxm−am=xn−anx−a⋅x−axm−am=xn−anx−a÷xm−amx−a \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:

lim⁡x→axn−anxm−am=lim⁡x→axn−anx−a÷xm−amx−a=(ddx(xn)∣x=a)÷(ddx(xm)∣x=a)=[nan−1]÷[mam−1]=nman−m \begin{aligned} \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{aligned}

Pi Han Goh - 5 years, 10 months ago

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

lim⁡x→axn−anxm−am=lim⁡x→addx(xn−an)ddx(xm−am)=lim⁡x→anxn−1mxm−1\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}}

=lim⁡x→anmxn−m=nman−m\displaystyle =\lim_{x\to a}\frac{n}{m}x^{n-m}=\frac{n}{m}a^{n-m}.

Maggie Miller - 5 years, 10 months ago

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

Mohamed Ahmed Abd El-Fattah - 5 years, 10 months ago

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

Mohamed Ahmed Abd El-Fattah - 5 years, 10 months ago
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