# Limits and Indeterminate Forms

1. Feb 4, 2016

### darkchild

Is it true that every limit that takes on an indeterminate form can be evaluated?

Is it proper to say that a limit problem has a solution if the limit does not exist?

2. Feb 4, 2016

### Staff: Mentor

It depends on how you define "evaluated." If, when you attempt to evaluate a limit, you get an indeterminate form, there are techniques that you can use to either a) evaluate the limit (get a number), or b) say that the limit doesn't exist (which includes $\infty$ as the "value" of the limit).
We don't say that a limit problem "has a solution." Equations and inequalities have solutions. A limit can be a) a finite number, b) unbounded, or c) not exist at all.
$\lim_{x \to \infty} x^2$ doesn't exist, in the sense that it is unbounded. We can also say that $\lim_{x \to \infty} x^2 = \infty$. All this means is that $x^2$ grows large without bound as x gets large.
$\lim_{n \to \infty} (-1)^n$ doesn't exist, period, because it oscillates forever between the two values, 1 and -1.

3. Feb 4, 2016

### darkchild

Ok, then is it appropriate to say that every indeterminate form can be simplified?

4. Feb 4, 2016

### Staff: Mentor

Like I said, when you get an indeterminate form, there are techniques (such as L'Hopital's Rule or algebraic techniques) that you can use to evaluate the limit or say that it doesn't exist. I wouldn't call this "simplifying" the limit expression, though. L'Hopital's Rule applies only to the $[\frac{-\infty}{\infty}]$ and $[\frac 0 0]$ indeterminate forms. Other indeterminate forms, such as $[1^{\infty}]$, require different techniques.

5. Feb 4, 2016

### pwsnafu

It's worth pointing out that even L'Hopital can't be applied to every $\frac\infty\infty$ indeterminate form either. For example $\frac{x+\sin(x)}{x}$ as $x\to\infty$.