Well that's it.
$$ \lim_{x\to \space-\infty} x^4+x^5 $$
How do I go about dealing with the $\infty-\infty$.
Thanks!
On
While $\infty-\infty$ is indeed indeterminate, $x^5$ grows in absolute value far faster than $x^4,$ so the limit is simply $\lim_{x\rightarrow -\infty}x^5=-\infty.$
On
A little more formal:
$\lim _{x \rightarrow -\infty} p(x) := x^4+ x^5 =$
$ x^5(1/x +1)$
Let $y:= - x$; and $P(y):= p(-y)$.
We now consider
$\lim_{ y \rightarrow \infty} P(y):$
$P(y)= -y^5(1-1/y) \lt $
$-y^5 (1/2) \lt -y/2,$ for $y > 2$.
$\lim_{y \rightarrow \infty} P(y):$
For every $2M \in \mathbb{R^+}$, $2M \gt 2$,
and $y \gt 2M$:
$P(y) \lt -y/2 \lt -M.$
Try factoring out the "dominant term" -- which in this case is $x^5$. By rewriting this as $$\lim_{x \to -\infty} x^5 \left( \frac 1 x + 1 \right)$$ you will have a product of terms that will be more manageable than the $\infty - \infty$ case.