Well that's it.

$$ \lim_{x\to \space-\infty} x^4+x^5 $$

How do I go about dealing with the $\infty-\infty$.


  • 3
    $\begingroup$ Write it as $x^4(1+x)$. $\endgroup$ – Thomas Andrews Oct 4 '17 at 1:36

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.

  • $\begingroup$ Thanks! Appreciate it! $\endgroup$ – undefined Oct 4 '17 at 1:45

While $\infty-\infty$ is indeed indeterminate, $x^5$ grows in absolute value far faster than $x^4,$ so the limit is simply $\lim_{x\rightarrow -\infty}x^5=-\infty.$

  • $\begingroup$ That's a nice analytical approach, I actually haven't studied it that way. Thanks! $\endgroup$ – undefined Oct 4 '17 at 1:46

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 $y \gt 2M$:

$P(y) \lt -y/2 \lt -M.$


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