trigonometry limits?

Question

what is the limit of sin (1/x) as x approaches infinity?
is it infinity or is it zero?

what is the limit of 2cos(2x) as x approaches infinity?
is it infinity or zero?

please explain this..thanks

Answer 1

the sin(1/x) as x approaches infinity is 0 since sin(0) = 0

The limit doesn't exists on cos function because it oscillates.

Answer 2

sin and cos functions only ever take values between -1 and 1, so neither of these functions could ever go to infinity.

The first one will be easier to talk about if we call u = 1/x. Then, as x approaches infinity, 1/x approaches zero, or in other words u approaches zero. And as u approaches zero, sin u approaches... well you should know that. This is your answer, because sin u = sin (1/x).

The second one, cos(2x) keeps oscillating between -1 and 1 as x approaches infinity. So 2cos(2x) oscillates between -2 and 2, every time x increases by pi. Thus there is no limit.

Now if you divided each of these functions by x, that would make the problems more interesting...

Answer 3

For the first question, lim[x --> infinity] sin (1/x) = 0. It doesn't matter here if x tends to plus or minus infinity, so for sake of the argument, assume that x increases towards the positive infinity.

Assume that E > 0 arbitrary small. If we can find an M such that for all x > M, |sin (1/x)| < E, we have proven the validity of the limit.

sin (1/x) < E <=> 1/x < sin^{-1} (E) <=> x > 1/[sin^{-1} (E)],

so if we set M = 1/[sin^{-1} (E)], |sin (1/x)| < E will hold for all x > M.

* sin^{-1} is the inverse of the sine function.

The second expression, 2 cos (2x), has no limit for x approaching infinity since it continues to oscillate between -2 and +2 with a period of pi.

Answer 4

As x approaches infinity, 1/x approaches zero, thus sin (1/x) approaches sin (0) = 0

The limit of 2 cos(2x) has no specific limit as x approaches infinity. It oscillates between -2 and +2 indefinitely.

Answer 5

As x aproaches infinity, (1/x) goes to zero.
What is the sine of zero?

As for 2cos(2x), that's indeterminate. Cosine of anything will be something between -1 and 1. Two times that will be something between -2 and 2. However, as x goes to infinity, cos(2x) doesn't converge to any particular value, since the wave function keeps oscillating between -2 and 2 forever.

Answer 6

For x -> 0 :
1) This limit doesnt exist the function oscilates between +1 and -1

2)same as 1