need help finding a limit

**cowboys111** · May 19th 2008, 02:01 PM

Hi, this limit is confusing me can someone please explain this one to me:
Limit as k goes to infinity k^2*sin^2(1/k).
Thanks for the help

**skyeam** · May 19th 2008, 02:05 PM

k² = infinity * infinity = infinity
1/k = 0, because divide through infinity gives 0

sin (0) = 0

sin² (o) = 0² = 0

0 * k² = 0 * infinity = NOT DEFINED

**Moo** · May 19th 2008, 02:07 PM

Hello,

Originally Posted by cowboys111

Hi, this limit is confusing me can someone please explain this one to me:
Limit as k goes to infinity k^2*sin^2(1/k).
Thanks for the help

Substitute : u=1/k.

When k goes to infinity, u goes to 0.

$\lim_{k \to \infty} k^2 \sin^2(1/k)=\lim_{u \to 0} \frac{\sin^2 u}{u^2}$

$=\lim_{u \to 0} \left(\frac{\sin u}{u}\right)^2=\lim_{u \to 0} \left(\frac{\sin u-\sin 0}{u-0}\right)^2$

But $\lim_{u \to 0} \frac{\sin u-\sin 0}{u-0}$ is the value of the derivative of $\sin u$ when u equals 0.
The derivative of $\sin u$ is $\cos u$

Therefore, $\lim_{u \to 0} \left(\frac{\sin u-\sin 0}{u-0}\right)^2=(\cos 0)^2=1$

**Mathstud28** · May 19th 2008, 02:20 PM

Originally Posted by cowboys111

Hi, this limit is confusing me can someone please explain this one to me:
Limit as k goes to infinity k^2*sin^2(1/k).
Thanks for the help

Alternately

$L=\lim_{n\to\infty}n^2\sin^2\bigg(\frac{1}{n}\bigg )=\lim_{n\to\infty}\bigg[n\sin\bigg(\frac{1}{n}\bigg)\bigg]^2$

So then

$\sqrt{L}=\lim_{n\to\infty}n\sin\bigg(\frac{1}{n}\b igg)=\lim_{n\to\infty}\frac{\sin\bigg(\frac{1}{n}\ bigg)}{\frac{1}{n}}$

Note this is an indetermiate form so we can apply L'hopitals

so $\lim_{n\to\infty}\frac{\sin\bigg(\frac{1}{n}\bigg) }{\frac{1}{n}}=\lim_{n\to\infty}\frac{\frac{-1}{n^2}\cos\bigg(\frac{1}{n}\bigg)}{\frac{-1}{n^2}}=\lim_{n\to\infty}\cos\bigg(\frac{1}{n}\bi gg)=1$

So $\sqrt{L}=1\Rightarrow{L=1^2=1}$

**Moo** · May 19th 2008, 02:30 PM

Note this is an indetermiate form so we can apply L'hopitals

I was waiting for this, I knew you would reply by using it !

**Mathstud28** · May 19th 2008, 02:36 PM

Originally Posted by Moo

I was waiting for this, I knew you would reply by using it !

...well I was going to do it your way but mid-type I saw you did it! So blame yourself

**Krizalid** · May 19th 2008, 05:24 PM

But having $n\sin \frac{1}{n}$ as $n\to\infty$ we only need a substitution and we'll get a well known limit.

**Mathstud28** · May 19th 2008, 05:37 PM

Originally Posted by Krizalid

But having $n\sin \frac{1}{n}$ as $n\to\infty$ we only need a substitution and we'll get a well known limit.

If your statement is directed at me, I usually try to do the weird abstract method that is easier but harder to see. But when I am beaten to it I show the more commonly known method such as L'hopital's. Substitution in limits can be a foreign concept to most high schoolers/college students so I showed this method to highlight how the limit is probably meant to be solved in their class or hwo to apply a known theorem.

**Krizalid** · May 19th 2008, 05:39 PM

It's not, actually, directed to ya, I said it as another way to tackle the problem.

And as for the "foreign" concept, that's false, it depends how one does teach such technique.

Thread: need help finding a limit

LinkBack

Thread Tools

Display

need help finding a limit

Similar Math Help Forum Discussions

Finding the limit

Finding limit of this function, using Limit rules

[SOLVED] Finding f(x) and g(x) - limit

Finding a limit. Finding Maclaurin series.

help finding the following limit:

Search Tags