If $$\lim_{x \to \infty}\frac{a(2x^3x^2)+b(x^3+5x^21)c(3x^3+x^2)}{a(5x^4x)bx^4+c(4x^4+1)+2x^2+5x}=1$$ 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
Consider $$ \lim_{x \to \infty}\frac{a(2x^3x^2)+b(x^3+5x^21)c(3x^3+x^2)}{a(5x^4x)bx^4+c(4x^4+1)+2x^2+5x}=1. $$ rewrite this as $$ \lim_{x \to \infty}\frac{x^3(2a+b3c)+x^2(a+5bc)b}{x^4(5ab+4c)+2x^2+x(a+5)+c}=1. $$ 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 $$5ab+4c=0.$$ Now same thing should happen in the numerator, the coefficient of $x^3$ term has to be zero. So we have $$2a+b3c=0.$$ This leads us to $$ \lim_{x \to \infty}\frac{x^2(a+5bc)b}{2x^2+x(a+5)+c}=1. $$ 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
$$ \frac{a+5bc}{2}=1. $$ Now you have three equations in $a,b$ and $c$ to solve.
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2A very well explained answer. +1 – Paramanand Singh Feb 26 '15 at 11:20
hint: $2a+b3c = 0 = 5ab+4c, a + 5b  c = 2$, you can solve for $a,b,c$.
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