Question

lim Cos(x^5)/x
x->-inf

Answer 1

The cosine function is always bounded between -1 and 1. Thus the maximum and minimum situations are as follows:\[\lim_{x\rightarrow -\infty}\dfrac{-1}{x}\le \lim_{x\rightarrow -\infty}\dfrac{\cos(x^5)}{x} \le \lim_{x\rightarrow -\infty}\dfrac{1}{x}\]We can evaluate the left and right hand side limits easily:\[0\le \lim_{x\rightarrow -\infty}\dfrac{\cos(x^5)}{x} \le 0\]Therefore, by the squeeze theorem:\[\lim_{x\rightarrow -\infty}\dfrac{\cos(x^5)}{x} = 0\]

Answer 2

okay so as x goes to -inf does it make a difference

Answer 3

What do you mean?

Answer 4

as x goes to negative infinity instead of positive infinity does it make a difference because cosine is confined to 1 and -1

Answer 5

They will both be the same, and can be proven via the same method that I used above.

Answer 6

okay so if it was 5cos(x^5) the cosine will be confined to -5 and 5 right

Answer 7

Also take note that the function is odd, so if it converges to some number going to negative infinity, it is going to converge to negative that number going to positive infinity. Since the number it converges to is zero, the number it converges to on the other side will also be zero.

Answer 8

thanks

Answer 9

Do you use squeeze theorem for all limit at infinity question or just rational function for example 3x^7+x^2

Answer 10

I used it for this situation because a component of the function we were taking the limit of had a definite upper and lower bound. You can test to see the behavior of the bounds and if they both converge to the same thing, the squeeze theorem supports that anything in between will converge to that value too.

Answer 11

Alright I will post the other question

Answer 12

very pretty yake :)