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
Hello,
Substitute : u=1/k.
When k goes to infinity, u goes to 0.
$\displaystyle \lim_{k \to \infty} k^2 \sin^2(1/k)=\lim_{u \to 0} \frac{\sin^2 u}{u^2}$
$\displaystyle =\lim_{u \to 0} \left(\frac{\sin u}{u}\right)^2=\lim_{u \to 0} \left(\frac{\sin u-\sin 0}{u-0}\right)^2$
But $\displaystyle \lim_{u \to 0} \frac{\sin u-\sin 0}{u-0}$ is the value of the derivative of $\displaystyle \sin u$ when u equals 0.
The derivative of $\displaystyle \sin u$ is $\displaystyle \cos u$
Therefore, $\displaystyle \lim_{u \to 0} \left(\frac{\sin u-\sin 0}{u-0}\right)^2=(\cos 0)^2=1$
Alternately
$\displaystyle L=\lim_{n\to\infty}n^2\sin^2\bigg(\frac{1}{n}\bigg )=\lim_{n\to\infty}\bigg[n\sin\bigg(\frac{1}{n}\bigg)\bigg]^2$
So then
$\displaystyle \sqrt{L}=\lim_{n\to\infty}n\sin\bigg(\frac{1}{n}\b igg)=\lim_{n\to\infty}\frac{\sin\bigg(\frac{1}{n}\ bigg)}{\frac{1}{n}}$
Note this is an indetermiate form so we can apply L'hopitals
so $\displaystyle \lim_{n\to\infty}\frac{\sin\bigg(\frac{1}{n}\bigg) }{\frac{1}{n}}=\lim_{n\to\infty}\frac{\frac{-1}{n^2}\cos\bigg(\frac{1}{n}\bigg)}{\frac{-1}{n^2}}=\lim_{n\to\infty}\cos\bigg(\frac{1}{n}\bi gg)=1$
So $\displaystyle \sqrt{L}=1\Rightarrow{L=1^2=1}$
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.