$\displaystyle \lim_{x \to \infty} x^6-3x^2+1$
How do you calculate this limit?
If you take the limit of a polynomial expression, it will shoot off to $\displaystyle +\infty$ or $\displaystyle -\infty$, depending on the end behaviour of the function.
In this case, consider the end behaviour of $\displaystyle x^6-3x^2+1$ to be $\displaystyle x^6$. Then it follows that $\displaystyle \lim_{x\to\infty}x^6-3x^2+1\sim\lim_{x\to\infty}x^6$. Since $\displaystyle x^6>0\,\forall\,x\in\mathbb{R},\, \lim_{x\to\infty}x^6=+\infty$.
Thus, $\displaystyle \lim_{x \to \infty} x^6-3x^2+1=+\infty$
Does this make sense?
Right, basically you base your value on the leading coefficient.
An easier way to explain it may be that:
$\displaystyle x^6$ will yield a "larger" infinity than $\displaystyle 3x^2$.
So when that is subtracted, the value is still $\displaystyle +\infty$.
The $\displaystyle 1$ on the end of the polynomial is irrelevant in this case.
Truthfully there is no such thing as a "larger" or "smaller" infinity, however this is an easier way to look at it.
Well basically, $\displaystyle \lim_{x \to \infty} x^6-3x^2+1=\lim_{x \to \infty} x^2(x^4-3)+1=(\infty)(\infty)+1=\infty$
But in general, you can use intuition on these limits. x^6 blows x^2 out of the water, no matter what the coefficient is on x^2, so -3x^2 doesn't matter as x gets huge (tends to infinity) and the plus 1 does not matter either when x is huge so in the long run, the polynomial acts exactly like x^6. and $\displaystyle \lim_{x \to \infty} x^6=\infty$
Hello,
A more straighforward way :
$\displaystyle x^6-3x^2+1=x^6\left(1-\frac{3}{x^4}+\frac{1}{x^6}\right)$
$\displaystyle \lim_{x\to\pm\infty} \frac{3}{x^4}=0$
$\displaystyle \lim_{x\to\pm\infty} \frac{1}{x^6}=0$
So $\displaystyle \lim_{x\to\pm\infty} x^6\left(1-\frac{3}{x^4}+\frac{1}{x^6}\right)=+\infty$
(because $\displaystyle x^6\geq 0$)