Let $\lim_{x\to 1}\frac{x^6+1+\frac{5}{3}(x^4+x^2)-\frac{8}{3}(x^5+x)}{x^5+x+6x^3-4x^4-4x^2}=\frac{p}{q}$,where $p,q\in N$

Question

Let $\lim_{x\to 1}\dfrac{x^6+1+\frac{5}{3}(x^4+x^2)-\frac{8}{3}(x^5+x)}{x^5+x+6x^3-4x^4-4x^2}=\frac{p}{q}$,where $p,q\in N$.Find the smallest value of $p+q$.

I applied L Hospital rule but it is not working here,limit is not finite after two differentiations,How should i solve it then?

Answer 1

Hint

It's not complicate to make five differentiation... moreover, all term of degree less than five will make $0$ after 5 differentiation, so you just have to differentiate $x^6-\frac{8}{3}x^5$ at the numerator and $x^5$ at the denominator.

Answer 2

Observe that $$ x^5+x+6x^3 -4x^4 -4x^2 = x (x-1)^4 $$ and $$ x^6+1+\frac{5}{3}(x^4+x^2)-\frac{8}{3}(x^5+x)=\frac{1}{3}(x-1)^4(3x^2+4x+3). $$ Can you go on from here?

Answer 3

Perhaps easier, change all $x$ to $t+1$, and study the limit as $t\to 0$ (this is essentially equivalent to what @Siminore does, but it might be easier to find factors like $t$ instead of $x-1$). We get that $$ \frac{x^6+1+\frac{5}{3}(x^4+x^2)-\frac{8}{3}(x^5+x)}{x^5+x+6x^3-4x^4-4x^2} =\frac{\frac{10}{3}t^4+\frac{10}{3}t^5+t^6}{t^4+t^5}=\frac{10+10t+3t^2}{3+3t}. $$ I think it is clear from here what the limit is, as $t\to0$.