If $$\lim_{x \to \infty}\frac{a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2)}{a(5x^4-x)-bx^4+c(4x^4+1)+2x^2+5x}=1$$ then find the value of $a+b+c$. I have done such problems with $x$ tending to some finite amount. Infinity is creating some problems.


2 Answers 2


Consider $$ \lim_{x \to \infty}\frac{a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2)}{a(5x^4-x)-bx^4+c(4x^4+1)+2x^2+5x}=1. $$ rewrite this as $$ \lim_{x \to \infty}\frac{x^3(2a+b-3c)+x^2(-a+5b-c)-b}{x^4(5a-b+4c)+2x^2+x(-a+5)+c}=1. $$ As long as the $x^4$ term is in the denominator the limit cannot be $1$, so the coefficient of that has to be $0$. So we get $$5a-b+4c=0.$$ Now same thing should happen in the numerator, the coefficient of $x^3$ term has to be zero. So we have $$2a+b-3c=0.$$ This leads us to $$ \lim_{x \to \infty}\frac{x^2(-a+5b-c)-b}{2x^2+x(-a+5)+c}=1. $$ Now divide both numerator and denominator by $x^2$ and use the fact that $1/x$ and $1/x^2 \to 0$ as $x \to \infty$ to get

$$ \frac{-a+5b-c}{2}=1. $$ Now you have three equations in $a,b$ and $c$ to solve.

  • 2
    $\begingroup$ A very well explained answer. +1 $\endgroup$
    – Paramanand Singh
    Feb 26, 2015 at 11:20

hint: $2a+b-3c = 0 = 5a-b+4c, -a + 5b - c = 2$, you can solve for $a,b,c$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy