How do you use the Squeeze Theorem to find #lim ( sin (2pi/x))# as x approaches zero?

Question

How do you use the Squeeze Theorem to find #lim  ( sin (2pi/x))# as x approaches zero?

Emily Ammerman · Accepted Answer

# lim_(xrarr0) sin((2pi)/x)# Does Not Exist

Here is the key idea. However close to #0# we start, it will happen that as #x rarr0#, we get infinitely many values of #(2pi)/x# that are coterminal with #pi/2#, so they have #sin((2pi)/x)=1#. (Specifically, for every #x=4/(1+4n# with #n# an integer.) We also get infinitely many values of #(2pi)/x# that are coterminal with #(3pi)/2#, so they have #sin((2pi)/x)=-1#. (Specifically, for every #x=4/(3+4n# with #n# an integer.) #sin((2pi)/x)# cannot be getting closer and closer to some number #L#, because we will always have some #x#'s for which the distance is #abs(L-1)# and others for which the distance is #abs(L-(-1)) = abs(L+1)#. These distances cannot both go to #0#. Applying the definition Remember the meaning: #lim_(xrarra)f(x)=L# if and only if For every #epsilon > 0#, there is a #delta > 0# for which for every #x# with #0 < abs(x-a) < delta#, we have #abs(f(x)-L) < epsilon#. To use the definition to show that a limit does not exist requires showing that there is an #epsilon> 0# for which, for every #delta > 0# there is an #x# with #0 < abs(x-a) < delta#, but #abs(f(x)-L) >= epsilon#. Given #L# (a proposed limit), choose #epsilon = 1#. Claim: for every #delta > 0# there is an #x# with #0 < abs(x-0) < delta#, but #abs(sin((2pi)/x)-L) >= epsilon# Proof of claim: For every #delta > 0#, there is an integer #n# for which #abs(4/(1+4n)) < delta# and #abs(4/(3+4n)) < delta#. Let #x_1 = 4/(1+4n)# and #x_2 = 4/(3+4n)# for such an #n#. Then #abs(f(x_1)-L) = abs(1-L) = abs (L-1)#. and #abs(f(x_2)-L) = abs(-1-L) = abs (L+1)#. Claim: we cannot have both #abs (L+1) < 1#. and #abs (L+1) < 1#. Suppose #abs (L+1) < 1#. Then #-1 < L+1 < 1#. But his implies that #-3 < L - 1 <-1#, so #abs(L-1) >=1# The proof is now complete.

Charles Anctil · Answer

undefined To use the Squeeze Theorem to find the limit of sin(2π/x) as x approaches zero, we need to find two functions that "squeeze" the given function and have the same limit as x approaches zero.

First, we observe that -1 ≤ sin(2π/x) ≤ 1 for all x ≠ 0.

Next, we consider two functions:

1. The constant function f(x) = 1, which is always greater than or equal to sin(2π/x) for all x ≠ 0.

2. The constant function g(x) = -1, which is always less than or equal to sin(2π/x) for all x ≠ 0.

Since both f(x) and g(x) have the same limit as x approaches zero (which is 1), and sin(2π/x) is always between f(x) and g(x), we can conclude that the limit of sin(2π/x) as x approaches zero is also 1.