Indeterminant?

[icystrike]

1. The problem statement, all variables and given/known data

http://img190.imageshack.us/img190/3204/84956253.jpg

This is not a homework.. I am wondering if it should really be zero or indeterminant form and i need a explaination(dont hesitate to quote from theorems)

2. Relevant equations



3. The attempt at a solution

[HallsofIvy]

The first two, \lim_{k\to\infty} 0= 0 and \lim_{k\to\infty} 0\cdot k= 0 are correct- they are both the limit of the sequence 0, 0, 0, ...

The third one is not "inderminant"- it does not exist because \lim_{k\to\infty} k does not exist.

[icystrike]

Do pardon me.. why \lim_{k\to\infty} k does not exist?

(I'm suspecting the determinant form of the indeterminant is 0)

[Mark44]

\lim_{k \to \infty} k = \infty

The limit doesn't actually exist, since \infty isn't a finite number. All this says is that as k gets large without bound, then (obviously) k gets large without bound. This is not one of the indeterminate forms (no such word as indeterminant) such as the following:

\left[\frac{\infty}{\infty}\right]
\left[\infty - \infty\right]\\
\left[\frac{0}{0}\right]\\
\left[1^{\infty}\right]

[icystrike]

\lim_{k \to \infty} k = \infty

The limit doesn't actually exist, since \infty isn't a finite number. All this says is that as k gets large without bound, then (obviously) k gets large without bound. This is not one of the indeterminate forms (no such word as indeterminant) such as the following:

\left[\frac{\infty}{\infty}\right]
\left[\infty - \infty\right]\\
\left[\frac{0}{0}\right]\\
\left[1^{\infty}\right]


However, we can always change into http://physicsforums.com/attachment.php?attachmentid=29483&stc=1&d=1288370729 which is indeed 0/0

[micromass]

The rule

\lim_{n\rightarrow +\infty}{x_ny_n}=\lim_{n\rightarrow +\infty}{x_n}\lim_{n\rightarrow +\infty}{y_n}

which you use, does not always hold. It only holds if the two limits on the right-hand side exists. And this is not the case here...

[Mark44]

However, we can always change into http://physicsforums.com/attachment.php?attachmentid=29483&stc=1&d=1288370729 which is indeed 0/0
Indeed it is not. The denominator is approaching 0, but the numerator is 1.

[icystrike]

There is a 0 outside the limit won't it make it 0/0?
Secondly, if limit of k as k tends to infinity does not exist, how about limit of k/k^2 as k tends to inifinity? ( since you can break up the limit to (lim k)/(lim k^2)

[Mark44]

That font is so small in that thumbnail that I took it to be 0 - the rest.

In any case, I don't see the point in going to complicated expressions just to represent
\lim_{k \to \infty} k = \infty.

In one sense, which is what HallsOfIvy was saying, the limit doesn't exist, since infinity isn't a value in the real number system. To say that the limit is infinity just means that the value of k gets larger and larger as k gets larger and larger.

\lim_{k \to \infty} \frac{k}{k^2} = \lim_{k \to \infty} \frac{1}{k} = 0

The first expression is one of the indeterminate forms I mentioned earlier in this thread. It can be simplified to the second expression above, which has a limit of 0.