# Thread: need help finding a limit

1. ## need help finding a limit

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

2. kČ = infinity * infinity = infinity
1/k = 0, because divide through infinity gives 0

sin (0) = 0

sinČ (o) = 0Č = 0

0 * kČ = 0 * infinity = NOT DEFINED

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

4. 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 $\sqrt{L}=1\Rightarrow{L=1^2=1}$

5. 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 !

6. 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

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

8. 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.

9. 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.