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If $\space a,b \in \mathbb{N}$ ,then calculate the limit $$\lim_{x\to1} \space\space \space \dfrac{a}{1-x^{a}}-\dfrac{b}{1-x^{b}}$$ and hence find $$ \lim_{x\to1} \space\space \space \dfrac{40}{1-x^{40}}-\dfrac{30}{1-x^{30}}$$ My approach: simplified $\lim_{x\to1} \space\space \space \dfrac{40}{1-x^{40}}-\dfrac{30}{1-x^{30}}$
to $\lim_{x\to1} \space\space \space \dfrac{5}{1-x^{5}}-\dfrac{15}{1-x^{15}}$

Unable to solve further. and also not able to generalize the result.

Any help would be appreciated .

THANKS!

J.G.
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user 34152
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1 Answers1

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Using L'Hôpital's rule twice, we seek$$\begin{align}\lim_{x\to1}\frac{a-b+bx^a-ax^b}{1-x^a-x^b+x^{a+b}}&=\lim_{x\to1}\frac{abx^{a-1}-abx^{b-1}}{-ax^{a-1}-bx^{b-1}+(a+b)x^{a+b-1}}\\&=\lim_{x\to1}\frac{ab(a-1)x^{a-2}-ab(b-1)x^{b-2}}{-a(a-1)x^{a-2}-b(b-1)x^{b-2}+(a+b)(a+b-1)x^{a+b-2}}\\&=\frac{ab(a-b)}{-a(a-1)-b(b-1)+(a+b)(a+b-1)}\\&=\frac{a-b}{2}.\end{align}$$While the limit you seek is $5$, the limit you thought was equivalent is in fact $-5$.

J.G.
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  • A very good method would be to replace x by $\frac{1}{x}$ in the original question. –  Dec 23 '20 at 10:23