$$\lim_{x\to \infty} \frac{6x^4+4}{(x^2-2)(6x^2-1)}$$
Based on the way the function behaves I can say it will reach 1, but I can't seem to be able to calculate the limit. Thanks for the help!
$\endgroup$ 35 Answers
$\begingroup$Multiply the numerator and denominator by $1/x^4$, then take the limit.
$\endgroup$ $\begingroup$Hint. Expand the denominator and use dominant terms.
$\endgroup$ 0 $\begingroup$Set $1/x=y$ to get
$$\lim_{y\to0}\dfrac{(6+4y^4)y^4}{(1-2y^2)(6-y^2)y^4}$$
Now cancel out $y$ as $y\to0$ as $y\ne0$
$\endgroup$ $\begingroup$Notice, $$\lim_{x\to \infty}\frac{6x^4+4}{(x^2-2)(6x^2-1)}$$ $$=\lim_{x\to \infty}\frac{6x^4+4}{x^2\left(1-\frac{2}{x^2}\right)x^2\left(6-\frac{1}{x^2}\right)}$$ $$=\lim_{x\to \infty}\frac{1}{x^4}\frac{6x^4+4}{\left(1-\frac{2}{x^2}\right)\left(6-\frac{1}{x^2}\right)}$$ $$=\lim_{x\to \infty}\frac{6+\frac{4}{x^4}}{\left(1-\frac{2}{x^2}\right)\left(6-\frac{1}{x^2}\right)}$$ $$=\frac{6+0}{\left(1-0\right)\left(6-0\right)}$$ $$=\frac{6}{6}=\color{red}{1}$$Alternative Method$$\lim_{x\to \infty}\frac{6x^4+4}{(x^2-2)(6x^2-1)}$$ $$=\lim_{x\to \infty}\frac{6x^4+4}{6x^4-13x^2+2}$$ $$=\lim_{x\to \infty}\frac{6+\frac{4}{x^4}}{6-\frac{13}{x^2}+\frac{2}{x^4}}$$ $$=\frac{6+0}{6+0}=1$$
$\endgroup$ $\begingroup$We factor the numerator and denominator by $x^4$. So, for $x$ big enough, we have $$ \frac{6x^4+4}{(x^2-2)(6x^2-1)} = \frac{6 + 4/x^4}{6 - 13/x^2 + 2/x^4}. $$ As $\lim_{x\to+\infty} 1/x^n = 0$ for all $n > 0$, we have $\lim_{x\to+\infty} 6 + 4/x^4 = 6$ and $\lim_{x\to+\infty} 6 - 13/x^2 + 2/x^4 = 6$. We conclude that $$ \lim_{x\to+\infty} \frac{6x^4 + 4}{(x^2-2)(6x^2-1)} = 1. $$
More generally, if $p(x) = a_n x^n + \dots + a_0$ and $q(x) = b_m x^m + \dots + b_0$ are polynomials of degree $n$ and $m$ respectively, then $$ \lim_{x\to\pm\infty} \frac{p(x)}{q(x)} = \lim_{x\to+\pm\infty} \frac{a_n}{b_m} x^{n-m}. $$ In other words, at infinity the behaviour of a rational function is given by the dominant terms of its denominator and numerator.
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