Limit of n²(1−cos(4.2/n)) using L'Hôpital's rule

[will_lansing]

Homework Statement

Find the limit of the sequence whose terms are given by $$a_{n}$$=$$n^{2}$$(1-cos$$\frac{4.2}{n}$$)

Homework Equations

The Attempt at a Solution

I now that $$n^{2}$$ goes to infinity so have to use l'hospital rule because you will have infinity*0 which is an indeterminate form, so rewrote as
$$a_{n}$$=$$\frac{(1-cos\frac{4.2}{n})}{\frac{1}{n^{2}}}$$

$$lim_{n\rightarrow\infty}$$ $$\frac{(1-cos\frac{4.2}{n})}{\frac{1}{n^{2}}}$$ $$\frac{}{}$$
= $$lim_{n\rightarrow\infty}\frac{$$ {$$\frac{-4.2sin\frac{4.2}{n}}{n^{2}}$$}/{$$\frac{-2}{n^{3}}$$}

won't the limit still be 0 i really don't understand any of this can anyone please help

 

[Dick]

Ok, so l'Hopital gives you the form 8.4*n*sin(4.2/n), which is still 0*infinity. Just apply l'Hopital again like you did the first time.

 

[will_lansing]

okay so if i do it again i get
8.4 $$lim_{n\rightarrow\infty}$$ $$\frac{sin\frac{4.2}{n}}{\frac{1}{n}}$$
so in the end i will get
8.4 $$lim_{n\rightarrow\infty}$$(-4.2cos(4.2/x))

did i do it right so far
so is the answer -35.28 i really don't understand how you are suppose to find the limit can someone please help

 

[Dick]

Ooops. I made a mistake. Sorry! It's not 8.4*n*sin(4.2/n). It's -2.1*n*sin(4.2/n). Change the 8.4 to -2.1. There are three minuses and the 4.2 is divided by 2. Now it's easy. As n->infinity what does the cos approach. There's a good reason not to give to explicit hints - because I make too many mistakes.

 

[will_lansing]

so as n approaches infinity cos approaches 0 right
so do you just multiply -2.1 by -4.2 to get the answer

 

[will_lansing]

yeah i got the right answer thanks so much. but i still think that finding limits for sequences are still hard.

 

[Dick]

Some are, some aren't. Your's is kind of in the middle. Good job, though. Sorry to confuse you.