Evaluate: $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$.
I've just started learning limits and calculus and this is an exercise problem from my textbook.
To solve the problem, I tried factorizing the numerator and denominator of the fraction. The numerator can be factorized as $(x-1)(x+1)(2x^2-1)$ and the numerator can be factorized as $x(6x^3+x^2-3x)$. So, we can rewrite the problem as follows: $$\lim\limits_{x\to\infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}=\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}$$ But this doesn't help as there is no common factor in the numerator and denominator. I've also tried the following: $$\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}=\lim\limits_{x\to\infty}\frac{x-1}{x}\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}=1\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}$$ Here I used $\frac{x-1} x=1$ as $x$ approaches infinity. Yet this does not help.
The answer is $\frac 1 3$ in the book but the book does not include the solution.
So, how to solve the problem?
Best Answer
Essentially, divide the numerator and denominator by the largest power of numerator:$$\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x} = \lim\limits_{x\to \infty}\frac {\frac{2x^4-3x^2+1}{x^4}}{\frac{6x^4+x^3-3x}{x^4}} = \lim\limits_{x\to \infty}\frac {2-\frac{3}{x^2} + \frac{1}{x^4}}{6 + \frac{1}{x} - \frac{3}{x^3}} = \frac{2 - 0 + 0}{6 + 0 - 0}= \frac13$$