Evaluate the limit: $\lim_{x\to \infty} \frac{(2x^2 +1)^2}{(x-1)^2(x^2+x)}$

Question

Evaluate the limit:

$$\lim_{x\to\infty} \frac{(2x^2 +1)^2}{(x-1)^2(x^2+x)}$$

The answer is 4 and I don't understand why, but why can't I just do something like:$$\frac{(\infty)}{(\infty)(\infty)} = \infty$$

Answer 1

\begin{align} \lim_{x\to\infty} \frac{(2x^2 +1)^2}{(x-1)^2(x^2+x)} &=\lim_{x \to \infty}\frac{4x^4+c_1x^3+c_2x^2+c_3x+c_4}{x^4+c_5x^3+c_6x^2+c_7x+c_8}\\ &=\lim_{x \to \infty}\frac{4+c_1x^{-1}+c_2x^{-2}+c_3x^{-3}+c_4x^{-4}}{1+c_5x^{-1}+c_6x^{-2}+c_7x^{-3}+c_8x^{x-4}}\\ &=4 \end{align} Except for terms of the highest power in both numerator and denominator, all terms $\to 0$ as $x \to \infty$

Answer 2

In mathematics one does not presume something is valid merely because one doesn't know a reason why it's not. And I don't know any reason why you would take $\dfrac{\infty}{\infty\cdot\infty}$ to be $\infty$.

A major reason for having the concept of a limit is "indeterminate forms" like $0/0$ and $\infty/\infty$, especially $0/0$. The derivative is defined as the limit of a quotient in which the numerator and denominator both approach $0$. The limit can be any number or $\infty$ or $-\infty$; it depends on which functions are approaching $0$.

$$ \frac{(2x^2 +1)^2}{(x-1)^2(x^2+x)} = \frac{4x^4+ 4x^2 + 1}{x^4 - x^3-x^2+ x} = \frac{4 + \dfrac4{x^2} + \dfrac{1}{x^4}}{1 - \dfrac1x-\dfrac1{x^2}+\dfrac1{x^3}}. $$ All of the fractions in the numerator and in the denominator in that last expression approach $0$ as $x$ approaches $\infty$.

Answer 3

$$ \lim_{x \to \infty}\dfrac{(2x^2 + 1)^2}{(x-1)^2(x^2 + x)} = \lim_{x \to \infty}\dfrac{\frac{(2x^2 + 1)^2}{(x^2)^2}}{\frac{(x-1)^2}{x^2}\frac{(x^2 + x)}{x^2}} = \lim_{x \to \infty} \frac{\biggl(2 +\frac{1}{x^2}\biggr)^2}{\biggl(1 - \frac{1}{x^2}\biggr)\biggl(1 + \frac{1}{x}\biggr)} = 4 $$

Answer 4

If I may suggest, let us consider the most general case of the limit, when $x \to \infty$, of the ratio of two polynomials $$F=\frac{P(x)}{Q(x)}=\frac{\sum _{i=0}^m a_i x^i}{\sum _{i=0}^n b_i x^i}$$ Since we look for large values of $x$, let us divide each term of the numerator by $x^m$ (its largest power) and each term of the denominator by $x^n$ (its largest power).

When $x$ is very large, we can ignore all terms which, after the divisions, show negative powers of $x$. So, basically, we obtain, for large values of $x$ $$F \simeq \frac{a_m x^m}{b_n x^n}= \frac{a_m }{b_n}x^{m-n}$$ and from here come the results

• If $m=n$, the limit of $F$ is just $\frac{a_m }{b_n}$
• If $m \gt n$, the limit is $+\infty$ if $\frac{a_m }{b_n} \gt 0$ and $-\infty$ if $\frac{a_m }{b_n} \lt 0$
• If $m \lt n$, the limit is $0$

Looking at your specific problem, without any development, you could see that the largest term of the numerator is $(2x^2)^2=4x^4$ and that the largest term of the denominator is $x^2 x^2=x^4$. So, ...