This is a question from my calculus text. It says $$\lim\limits_{x\to1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right)$$ where $p$ and $q$ are natural numbers. I know this is an infinity-infinity indeterminant form which can be converted to a $0/0$ form. I tried substituting $x=1+h$ where $h\to0$. But it is not working. What should I do?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
Sign up
Sign up
Here's how it works:
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
|
|
Considering $$A=\frac{p}{1-x^p}-\frac{q}{1-x^q}$$ let $x=1+h$ as you did; so $$A=\frac{p}{1-(1+h)^p}-\frac{q}{1-(1+h)^q}$$ Now, using the binomial theorem or Taylor series $$(1+h)^a=1+ah+\frac 12a(a-1)h^2+\cdots$$ $$1-(1+h)^a=-ah-\frac 12a(a-1)h^2+\cdots$$ Replace $$\frac{a}{1-(1+h)^a}=\frac{a}{-ah-\frac 12a(a-1)h^2+\cdots}=-\frac 1{h}\frac 1 {1+\frac 12 (a-1)h+\cdots}$$Now, long division gives $$-\frac 1{h}(1-\frac 12 (a-1)h+\cdots)$$ Now, replacing $a$ by $p$ and then $a$ by $q$ gives $$A\approx-\frac 1{h}(1-\frac 12 (p-1)h+\cdots)+\frac 1{h}(1-\frac 12 (q-1)h+\cdots)=\frac {p-q} 2+\cdots$$ Using one more term in the binomial expansion and doing the same would have given $$A\approx\frac {p-q} 2+\frac{q^2-p^2}{12}h$$ showing the limit and how it is approached. |
|||||||||
|
|
This simply adds a bit of rigor to Claude Leibovici's answer. Using Landau Big-O notation, $$ \begin{align} \frac{a}{1-x^a} &=\frac{a}{1-(1+h)^a}\\ &=\frac{a}{1-1-ah-\frac{a(a-1)}2h^2+O\!\left(h^3\right)}\\ &=-\frac1h\frac1{1+\frac{a-1}2h+O\!\left(h^2\right)}\\ &=-\frac1h\left(1-\frac{a-1}2h+O\!\left(h^2\right)\right)\\ &=-\frac1h+\frac{a-1}2+O(h)\\ &=-\frac1{x-1}+\frac{a-1}2+O(x-1)\tag{1} \end{align} $$ Applying $(1)$ for $a=p$ and $a=q$ gives that $$ \frac{p}{1-x^p}-\frac{q}{1-x^q}=\frac{p-q}2+O(x-1)\tag{2} $$ Therefore, $$ \lim_{x\to1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right)=\frac{p-q}2\tag{3} $$ |
|||
|
|
|
Note $$\frac{p}{1-x^p}-\frac{q}{1-x^q}=\frac{p(1-x^q)-q(1-x^p)}{(1-x^p)(1-x^q)}$$ which gives the indeterminate form. Simplify to get $$\frac{p-px^q-q+qx^p}{1-x^q-x^p+x^{p+q}}$$ and since $q,p$ are positive integers this is simply a polynomial where the detonator is of a strictly higher degree. |
|||||||||
|
|
The simplest approach is to use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ We have \begin{align} L &= \lim_{x \to 1}\left(\frac{p}{1 - x^{p}} - \frac{q}{1 - x^{q}}\right)\notag\\ &= \lim_{x \to 1}\frac{p - px^{q} - q + qx^{p}}{(1 - x^{p})(1 - x^{q})}\notag\\ &= \lim_{x \to 1}\dfrac{p - px^{q} - q + qx^{p}}{\dfrac{(1 - x^{p})(1 - x^{q})}{(1 - x)(1 - x)}\cdot(1 - x)^{2}}\notag\\ &= \frac{1}{pq}\lim_{x \to 1}\frac{p - q + qx^{p} - px^{q}}{(1 - x)^{2}}\notag\\ &= \frac{1}{pq}\lim_{h \to 0}\frac{p - q + q(1 + h)^{p} - p(1 + h)^{q}}{(1 - x)^{2}}\notag\\ &= \frac{1}{pq}\lim_{h \to 0}\dfrac{p - q + q\left(1 + ph + \dfrac{p(p - 1)}{2}h^{2} + \cdots\right) - p\left(1 + qh + \dfrac{q(q - 1)}{2}h^{2} + \cdots\right)}{h^{2}}\notag\\ &= \frac{1}{pq}\lim_{h \to 0}\dfrac{\dfrac{pq(p - q)}{2}\cdot h^{2} + \cdots}{h^{2}}\notag\\ &= \frac{p - q}{2}\notag\\ \end{align} The ellipsis ($\cdots$) used above indicates a finite number of terms (based on positive integers $p, q$) and each term is of the form $c\cdot h^{r}$ where $r > 2$. This is because of the standard binomial theorem for positive integer index. In case numbers $p, q$ are not positive integers then we use the general binomial theorem and the ellipsis $(\cdots)$ needs to be replaced with $o(h^{2})$. |
|||
|
|
|
Hint $\frac{a^n-1^n}{a-1}=na^{n-1}$ so you can multiply /divide by $1-x$ to get answer as $0$ |
|||||||||||||||||||||
|
