# Indeterminate Forms

#### orcsmoocher

Questions that don't tell you what you got wrong, O I hate thee. Apparently I have anywhere from 17-19 correct based on the grading scale put on this question.

The question:
For each of the following forms determine whether the corresponding limit type is indeterminate, always has a fixed finite value, or never has a fixed finite value.

I... really have no idea beyond the basic forms that were given during the lecture. Limits and infinity make me cry. So a few words to explain why which ones are wrong would be very kind

#### greg1313

Look at # 4.

It's a form of limit as x, y ---> oo of $$\displaystyle x^{-y} \ =$$

limit as x, y ---> oo of $$\displaystyle \dfrac{1}{x^y}$$

This looks to be 0 instead of "does not exist."

topsquark

#### orcsmoocher

Look at # 4.

It's a form of limit as x, y ---> oo of $$\displaystyle x^{-y} \ =$$

limit as x, y ---> oo of $$\displaystyle \dfrac{1}{x^y}$$

This looks to be 0 instead of "does not exist."
Dang, you're right! I admittedly didn't think to put the ones with negative exponents in the denominator to see "one over really small."

#### Debsta

MHF Helper
Dang, you're right! I admittedly didn't think to put the ones with negative exponents in the denominator to see "one over really small."
Do you mean "one over really big"?

#### Debsta

MHF Helper
#8 and 9 would be 1 rather than indeterminant.

topsquark

#### topsquark

Forum Staff
#8 and 9 would be 1 rather than indeterminant.
Here's something of interest. I've always learned that $$\displaystyle \lim_{n \to \infty} 1^n$$ is indeterminate. (I also looked it up and the wikipedia article agreed with me, anyway.) But W|A says it's 1. Strange.

-Dan

#### orcsmoocher

Do you mean "one over really big"?
Yes I was thinking " The Whole Thing Approaches 0" but my brain hadn't caught up yet. I promise I'm usually smarter than this haha

#### greg1313

Here's something of interest. I've always learned that $$\displaystyle \lim_{n \to \infty} 1^n$$ is indeterminate.
(I also looked it up and the wikipedia article agreed with me, anyway.) But W|A says it's 1. Strange.

-Dan
@ Debsta
@ topsquark

It (the form) is indeterminate. Please look at these examples after you put them into graphing/other calculators and/or computers:

limit as n ---> oo $$\displaystyle \$$ of $$\displaystyle \ \bigg(1 \ + \ \dfrac{1}{n^2}\bigg)^n$$

limit as n ---> oo $$\displaystyle \$$ of $$\displaystyle \ \bigg(1 \ + \ \dfrac{1}{n^2}\bigg)^{n^2}$$

limit as n ---> oo $$\displaystyle \$$ of $$\displaystyle \ \bigg(1 \ + \ \dfrac{1}{n^2}\bigg)^{n^3}$$

limit as n ---> oo $$\displaystyle \$$ of $$\displaystyle \ \bigg(1 \ - \ \dfrac{1}{n^2}\bigg)^n$$

limit as n ---> oo $$\displaystyle \$$ of $$\displaystyle \ \bigg(1 \ - \ \dfrac{1}{n^2}\bigg)^{n^2}$$

limit as n ---> oo $$\displaystyle \$$ of $$\displaystyle \ \bigg(1 \ - \ \dfrac{1}{n^2}\bigg)^{n^3}$$

There should be at least four different limit values from the above. I haven't checked them all.

(Edit) For similar reasons, the form in # 8 is also indeterminate.

Last edited:
topsquark and Idea

#### Idea

$$\displaystyle 1^n$$ is an indeterminate form e.g.

$$\displaystyle \left(1+\frac{1}{n}\right)^n\to e$$

and

$$\displaystyle \left(1-\frac{1}{n}\right)^n\to \frac{1}{e}$$

on the other hand

$$\displaystyle \lim_{n\to \infty } 1^n=1$$

topsquark

#### Jhevon

MHF Helper
Questions that don't tell you what you got wrong, O I hate thee. Apparently I have anywhere from 17-19 correct based on the grading scale put on this question.

The question:
For each of the following forms determine whether the corresponding limit type is indeterminate, always has a fixed finite value, or never has a fixed finite value.
View attachment 39613
I... really have no idea beyond the basic forms that were given during the lecture. Limits and infinity make me cry. So a few words to explain why which ones are wrong would be very kind
I've always thought of indeterminate forms this way (which, admittedly, you'll need a bit of number sense for, but it is arguably the kind of number sense you would develop anyway by doing a lot of math problems):

Indeterminate forms arise when you have competing, contradictory ideas clashing. For example, $$\displaystyle 0^0$$ is indeterminate because there's the idea that raising anything to the 0 power should give you 1 (except for 0 of course), while anything with base zero would normally be thought of as 0 (for positive powers anyway). There is a typical way of things unfolding. When two such ideas clash, indeterminate forms arise. So $$\displaystyle 1^\infty$$ is indeterminate because a base of 1 should give you 1, but thinking of exponential functions, they go off to infinity as the power goes to infinity. So one idea tells you the answer is 1, another tells you the answer should be infinite.

Again, there are nuances here, since $$\displaystyle a^x \to \infty$$ as $$\displaystyle x \to \infty$$ only when $$\displaystyle a>1$$, but hopefully you guys see what I'm saying...

topsquark