Recurrence relations with unknowns

Question

Let $$D_n = \frac{1}{b-a}((a^nb-ab^n)(a+b)+(b^n-a^n)(a^2+ab+b^2))$$

where $a≠b$.

Find $D_n$ if $a=b$ in terms of $a$ and $n$.

I had to solve a recurrence relation to end up with the above answer and the hint for this part is to let $b$ in my answer tend to $a$.

However, doesn't the function become undefined in such a case?

Any fast responses will be greatly appreciated, as I have to submit this in 30 minutes ):

Thanks in advance!

Answer 1

HINT

If you multiply out the numerator and simplify, you get $$ (a^nb-ab^n)(a+b)+(b^n-a^n)(a^2+ab+b^2) = b^{n+2} - a^{n+2} $$ Can you factor out $(a-b)$ or $(b-a)$ from that?

UPDATE

You have $$ \begin{split} (a^nb-ab^n)&(a+b)+(b^n-a^n)(a^2+ab+b^2) \\ &= \left[a^{n+1}b + a^nb^2 -a^2b^n - ab^{n+1}\right]\\ &+ \left[a^2b^n - a^{n+2} + ab^{n+1} - a^{n+1}b + b^{n+2} - a^nb^2 \right] \\ &= b^{n+2}-a^{n+2} \end{split} $$

Answer 2

Letting $b$ tend to $a$ does not make it undefined. You will get an indeterminate of the form $\frac 00$, but are taking the limit where $b \neq a$ but tends there. You need either to divide the factor $b-a$ out of the numerator analytically, which is easy for the second term but not for the first, or write $b=a+\epsilon$, let $\epsilon \to 0$ and use L'Hôpital's rule

Answer 3

The second term vanish, the first equals $$\dfrac{1}{b-a}(ab)(a^{n-1}-b^{n-1})(a+b)=-ab(a+b)(\sum_{i=0}^{n-2}a^ib^{n-2-i})$$

Answer 4

You can make it simple if you let $b=a(1+\epsilon)$. This would make $$D_n=a^{n+1}\frac{ \left((1+\epsilon )^{n+2}-1\right)}{\epsilon }$$Using the binomial expansion or equivalents $$(1+\epsilon )^{n+2}=1+(n+2)\epsilon+\cdots$$ Replacing $$D_n=(n+2)a^{n+1}$$