A simple problem that's baffling me is this:
what is
$$
\lim_{x\rightarrow\infty}{\frac{x^2-4}{-4x^2+x+2}}?
$$
Thanks for all the help. I have a test tomorrow and I'm reviewing a lot of old stuff.
$\endgroup$ 17 Answers
$\begingroup$$\lim_{x\rightarrow∞}{\frac{x^2-4}{-4x^2+x+2}} = \lim_{x\rightarrow∞}{\frac{1-\frac{4}{x^2}}{-4+\frac1x+\frac{2}{x^2}}}=-\frac14$
$\endgroup$ $\begingroup$Because polynomials are of the same degree, the limit is equal to the fraction of the coefficients at $x^2$, i.e. $-1/4$.
(This is a simple rule, without any calculations).
$\endgroup$ $\begingroup$As the degree of nominator and denominator are equal, hence at infinity, the limit will be the ratio of the coefficients of the greatest-degree polynomials at nominator and denominator, which in this example is $x^2$, hence $\lim=\frac{-1}{4}$
$\endgroup$ $\begingroup$I was told by a math teacher the following (very simplified!) shortcut re: lim x→∞
The answer is the coefficients of the highest exponent (in this case x^2) in the numerator or denominator. Answer: lim= -1/4 Another example: lim x→∞ (6x^4+3x^3-2x^2+8x-3)/(5x^4+1) Answer: lim = 6/5
This works unless you have a problem such as: lim x→∞ (5x+5)/(2x^2) In this case, your coefficients would be 0/2 because there are no x^2 in the numerator. Therefore, you have lim = 0/2 = 0
You also have to watch out for the following: If you have a problem such as lim x→∞ (x^2)/(x+5) In this case, your coefficients would be 1/0 as there are no x^2 in the denominator. Therefore, lim = 1/0 = ∞
I hope you can understand how I wrote this. I don't know how to write MathML code!
$\endgroup$ $\begingroup$Though all the other answers are correct, none seems to address the intuition behinf the method. I'll give it a try here.
Taking the limit of $f(x)=\frac{x^2-4}{-4x^2+x+2}$ at $\infty$ roughly means approximating the behavior of $f$ when $x$ is large. But, when $x$ is large, $-4x^2$ is very large, much larger than $x$ or $2$. So $-4x^2+x+2$ can be approximated by $-4x^2$ (think that $x=10^6$ or even greater number). And similarly, $x^2-4$ can be approximated by $x^2$. So, morally, $f(x)$ is close to $\frac{x^2}{-4x^2}=\frac{-1}{4}$ when $x$ is large. Therefore, we should expect that $\displaystyle{\lim_{x\to\infty}\frac{x^2-4}{-4x^2+x+2}=\frac{-1}{4} }$.
To prove it, the idea is to factor numerator and denominator by the "strongest term": we have
$$ \displaystyle{ f(x)=\frac{x^2(1-\frac{4}{x^2})}{-4x^2(1+-\frac{1}{4x}-\frac{1}{2x^2})} } $$
so we can canl out the $x^2$ and apply Limit Laws to obtain thay $\displaystyle{\lim_{x\to\infty}\frac{x^2-4}{-4x^2+x+2}=\frac{-1}{4} }$
(morally $f(x)=\frac{x^2(\textrm{something close to }1)}{-4x^2(\textrm{something close to }1)}=\frac{-1}{4}(\textrm{something close to }1)$).
This is useful to think that way since it can be applied to many other situations, for example the limit at $\infty$ of any rational function, or limits of the form $\displaystyle{\lim_{x\to\infty}\frac{a^x-b^x}{a^x+b^x}}$,..
$\endgroup$ $\begingroup$You can use L'Hopital's Rule, since after substitution, the numerator and the denominator are equal to $0$ or $+$ or $-$ $\infty$.
$\endgroup$ 3 $\begingroup$Simply divide numerator and denominator by $x^2$ then apply limit of infinity you got the answer,thanks and have a nice day with hard work...
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