Evaluating $\lim_{x\to 1}\frac{p \left(1 + \sum_{r=1}^{q-1}x^r \right)-q\left(1+\sum_{r=1}^{p-1}x^r\right)}{1-x} $ without L'Hospital's rule?

Question

The following limit is to be evaluated without using L'Hospital's rule.

$$\lim_{x\to 1}\frac{p \left(1 + \sum_{r=1}^{q-1}x^r \right)-q\left(1+\sum_{r=1}^{p-1}x^r\right)}{1-x} $$

(original problem image)

Despite many attempts (unsuccessful) I'm unable to evaluate the same. Any help would be highly appreciated.

The answer of the limit should be $\frac12pq(p-q)$.

NOTE: I’m a high school student.

Answer 1

$$\lim_{x\to 1}\frac{p((\sum_1^{q-1}x^r)+1)-q((\sum_1^{p-1}x^r)+1)}{1-x}$$

$$\lim_{x\to 1}\frac{p(x^{q-1}+x^{q-2}+.....+x+1)-q(x^{p-1}+x^{p-2}+.....+x+1)}{1-x}$$

$$\lim_{x\to 1}\frac{p\Bigg(\frac{(1-x)(x^{q-1}+x^{q-2}+.....+x+1)}{1-x}\Bigg)-q\Bigg(\frac{(1-x)(x^{p-1}+x^{p-2}+.....+x+1)}{1-x}\Bigg)}{1-x}$$

$$\lim_{x\to 1}\frac{p(1-x^q)-q(1-x^p)}{(1-x)^2}$$

$$Because ((1-x^n)=(1-x)(x^{n-1}+x^{n-2}+.....+x+1))$$

$$\lim_{x\to 1}\frac{p(1-(1-(1-x))^q)-q(1-(1-(1-x))^p)}{(1-x)^2}$$

Substituting (1-x)=t

$$\lim_{x\to 1}\frac{p(1-(1-t)^q)-q(1-(1-t)^p)}{t^2}$$

Now using Binomial Expansion

$$\lim_{x\to 1}\frac{\Bigg(p\Bigg(1-\Big(1-qt+\frac{(q)(q-1)t^2}{2}-\frac{(q)(q-1)(q-2)t^3}{3!}.....∞\Big)\Bigg)-q\Bigg(1-\Big(1-tp+\frac{(p)(p-1)t^2}{2}-\frac{(p)(p-1)(p-2)t^3}{3!}.....∞\Big)\Bigg)\Bigg)}{t^2}$$

$$\lim_{x\to 1}\frac{\Bigg(p\Big(qt-\frac{(q)(q-1)t^2}{2}+\frac{(q)(q-1)(q-2)t^3}{3!}+.....∞\Big)-q\Big(pt-\frac{(p)(p-1)t^2}{2}+\frac{(p)(p-1)(p-2)t^3}{3!}+.....∞\Big)\Bigg)}{t^2}$$

$$\lim_{x\to 1}\frac{(pqt-\frac{((pq)(q-1)t^2}{2}+\frac{(pq)(q-1)(q-2)t^3}{3!}+.....∞)-(pqt-\frac{(pq)(p-1)t^2}{2}+\frac{(pq)(p-1)(p-2)t^3}{3!}+.....∞)}{t^2}$$

$$\lim_{x\to 1}\Bigg(\Big(\frac{-(pq)(q-1)}{2}+\frac{(pq)(q-1)(q-2)t}{3!}+.....∞\Big)+\Big(\frac{(pq)(p-1)}{2}+\frac{(pq)(p-1)(p-2)t}{3!}+.....∞\Big)\Bigg)$$

$$\frac{(-pq)(q-1-p+1)}{2}$$

$$\frac{(pq)(p-q)}{2}$$

Answer 2

Remember first that $$\sum_{r=1}^n x^r=\frac{x \left(x^n-1\right)}{x-1}$$ So, computing the numerator, we have $$p\frac{x^q-1}{x-1}-q \frac{x^p-1}{x-1}=\frac{p \left(x^q-1\right)-q \left(x^p-1\right)}{x-1}$$ Now, as @Yves Daoust wrote in a deleted answer, let $x=1+y$ and apply the binomial theorem $$(1+y)^n=1+ n y+\frac {n(n-1)}2 y^2+\frac {n(n-1)(n-2)}6 y^3+\cdots$$

Just apply and simplify to get a nice result for the limit.

Answer 3

Note that $$ \begin{align} \lim_{x\to1}\frac{x^r-1}{x-1} &=\lim_{x\to1}\sum_{k=0}^{r-1}x^k\\ &=\sum_{k=0}^{r-1}1\\[6pt] &=r\tag1 \end{align} $$ Then $$ \begin{align} &\lim_{x\to1}\frac{p\left(1+\sum\limits_{r=1}^{q-1}x^r\right)-q\left(1+\sum\limits_{r=1}^{p-1}x^r\right)}{1-x}\\ &=\lim_{x\to1}\frac{p\sum\limits_{r=1}^{q-1}\left(x^r-1\right)-q\sum\limits_{r=1}^{p-1}\left(x^r-1\right)}{1-x}\tag2\\[6pt] &=\lim_{x\to1}q\sum\limits_{r=1}^{p-1}\frac{x^r-1}{x-1}-\lim_{x\to1}p\sum\limits_{r=1}^{q-1}\frac{x^r-1}{x-1}\tag3\\ &=q\sum\limits_{r=1}^{p-1}r-p\sum\limits_{r=1}^{q-1}r\tag4\\[6pt] &=q\frac{p(p-1)}2-p\frac{q(q-1)}2\tag5\\[9pt] &=\frac{pq(p-q)}2\tag6 \end{align} $$ Explanation:
$(2)$: subtract $pq$ from both sides of the difference in the numerator
$(3)$: apply the distributive property
$(4)$: apply $(1)$
$(5)$: sum the arithmetic sequence
$(6)$: simplify