Question

find the limit of x-cosx/x as x approches infinity

Answer 1

can you use L'Hopital's Rule yet?

Answer 2

Im not sure what that is, maybe

Answer 3

I don think the limit exists.

Answer 4

\[\frac{x-cos(x)}{x}\]?

Answer 5

yes, the limit of cos(x) as it goes to infinity is undefined isnt it?

Answer 6

would that make it just 1 because of x/x

Answer 7

well ok look at it this way
\[\frac{x-cos(x)}{x}\]divide top and bottom by x
\[\frac{\frac{x}{x}-\frac{cos(x)}{x}}{\frac{x}{x}}=\frac{1-\frac{cos(x)}{x}}{1}=1-\frac{cos(x)}{x}\]
now
\[-1\le cos(x)\le1\\and\\\frac{1}{x}\le \frac{cos(x)}{x}<=\frac{1}{x}\]
take the limit of this and we have 0
we we are left with \[\frac{1-0}{1}=1\]
s0
\[\lim_{x \rightarrow \infty }\frac{x-cos(x)}{x}=1\]

Answer 8

hmm not sure i understood all of that lol... but the answer is 1?

Answer 9

yes, the part with -1<=cos(x)<=1 is the squeeze theorem

Answer 10

if you just want the answer...
http://www.wolframalpha.com/input/?i=find+the+limit+of+%28x-cosx%29%2Fx+as+x+approches+infinity

Answer 11

alright ill look up the squeeze theorem, thanks

Answer 12

note we can hold the inequalities because we are considering x large, not negative.