Question

lim t-> 0 of sin(t)/t DNE right?

Answer 1

oh no

Answer 2

world famous limit
mother of all trig limits
it is one

Answer 3

alrighty. My book shows a matrix \[\left[\begin{matrix}t & t ^{-1} \\ 0 & e ^{t}\end{matrix}\right]\] and says that lim t-> 0 , this limit DNE. I assumed it was because of \[t ^{-1}\] would be a 1/0 which is undefined

Answer 4

so my instinct said that sint/t would be the same situation...

Answer 5

unless i am missing the point entirely, and some other element of the matrix is the reason the limit DNE

Answer 6

i have no idea what that means
you asked about
\[\lim_{t\to 0}\frac{\sin(t)}{t}\] which is one
not sure where the matrix came from

Answer 7

its just the example in the book of a matrix whos limit did not exist.

Answer 8

Are you familiar with power series @xartaan?

Answer 9

vaguely...

Answer 10

Well there are essentially two ways you can do this. The simplest way would be to start plugging in values closer and closer to 0. So try t=.1, .01, .001, .000001, etc... and see what you approach as you get nearer to 0.

The other method is to use the power series representation of sinx which says:

\[\LARGE \sin (x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]

So when you divide this by x you get an x term divided out of each:\[\LARGE \frac{\sin (x)}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...\]

Plugging in 0 will make all the infinite number of terms go away except the first one: \[\LARGE \lim_{x \rightarrow 0} \frac{\sin (x)}{x} = 1\]

Answer 11

I can explain either thing more if you're curious or something isn't clear enough.

Answer 12

i have a feeling that this is something entirely different

Answer 13

what does it mean "the limit of a matrix"?
perhaps it means
\[\lim_{n\to \infty}M^n\]

Answer 14

@satellite73 No, that was just an entirely separate example of a limit that didn't exist because we had the limit of 1/t. However the limit of 1/t is actually different than sin(t)/t even though both have a t in the denominator.

@xartaan If you're unsure as to why this limit would exist, in fact these are different because while in your example the limit is 1/0 in this example we actually have something called an indeterminant form which is 0/0 which may, and in this case definitely does, have a limit.

Answer 15

@satellite73 The limit of the matrix is just saying that as t approaches 0, it's the same as taking the limit of each individual component of the matrix. See if you try to compute each part, you get something like this:\[\Large \lim_{t \rightarrow 0}\left[\begin{matrix}t & t^{-1} \\ 0 & e^t\end{matrix}\right]=\left[\begin{matrix}0 & DNE \\ 0 & 1\end{matrix}\right]\]

It's just like finding the derivative of a matrix.

Answer 16

sorry internet went out. And also sorry that I wasn't clear earlier. Kainui has what I mean down though. So that matrix was given in the book as an example of how limits of a matrix work. They took that matrix and show that for a lim as t -> the limit of the matrix exists, yet as t-> the limit of the matrix does not. The book does not however, mention which element in the matrix is the reason for this,

Answer 17

My guess was that it was the t^(-1), which Kainui has shown above. So, I took this to mean that, because that element would be 1/0 and thus undefined, the same would hold true for sin(t)/t.

Answer 18

It has been a while since calculus and I feel a bit foolish not recognizing this was a case for indeterminant form. :( But I certainly appreciate the clarifying guys, thanks!

Answer 19

No problem, I'm always on the lookout for people who are trying to get better at calculus and are willing to help themselves. You seem to be doing both of those things, so keep it up haha!