finding limits question involving cos and sine?

Question

hi,

i would like to find out how to solve this problem

i would like to find out the following limits if they exist

lim cos (3 pi/x) as x approaches infinity

lim cos (4x^-1) as x approaches 0 from the right (x____>0+)

lim sin (3 pi x) as x approaches infinity

thanks for your help, it is much appreciated .

Answer 1

1)
lim ( cos(3pi/x) )
x -> infinity

As x approaches infinity, 3pi/x approaches 0, so this is no different than calculating cos(0).

cos(0) = 1

2)

lim ( cos( 4x^(-1) ) )
x -> 0+

Change x^(-1) into 1/x.

lim ( cos (4 (1/x) ) )
x -> 0+

lim ( cos(4/x) )
x -> 0+

As x approaches 0 from the right, 4/x approaches infinity, and the cosine of infinity is undetermined (because cosine oscillates). Therefore, the limit does not exist.

cos(0)

Which is equal to 1.

3)
lim ( sin (3pi(x)) )
x -> infinity

This limit does not exist, because as x gets very large, 3pi(x) gets very large, and the sine of a very large number is undetermined (because of the way sine oscillates).

Answer 2

lim (x->∞) cos (3 π/x)
As x->∞, 3π/x -> 0.
So lim (x->∞) cos (3 π/x)
= lim (x->∞) cos (0)
= lim (x->∞) 1
= 1.

lim (x->0+) cos (4x^-1)
- as x -> 0+, 4x^-1 -> +∞. So cos(4x^-1) will oscillate ever more rapidly between -1 and +1, so this limit does not exist.

lim (x->∞) sin (3 π x)
Similarly, at any point of this function, the next 2π values of x will see all values of the function from -1 to 1. So the limit does not exist.

Answer 3

For the 3pi/x-:

the cos of this will approach 1 (as cos 0 = 1)

Answer 4

cos(3pi/x)=cos(0)=1
cos(4x^-1)=cos(+inf)=undefined
sin(3pix)=sin(+inf)=undefined