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How do you use the Squeeze Theorem to find #limsin(1/x)# as x approaches zero?

Answer 1

The limit #lim_(x->0)sin(1/x)# does not exist. You can proove this taking two sequences such as #a_n=1/(2pin)# and #b_n=1/((2n+1)*pi)# and replacing into the value of x.Hence

#sin(1/(1/((2pin))))=sin(2*pi*n)=1 or -1#.Hence the values of the limits both sequences are different such limit does not exist

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Answer 2

Luke Alvarez

To use the Squeeze Theorem to find lim sin(1/x) as x approaches zero, we can establish two functions that "squeeze" the given function.

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

Next, we consider two additional functions:

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

Since f(x) ≥ sin(1/x) ≥ g(x) for all x ≠ 0, and both f(x) and g(x) approach 1 as x approaches zero, we can conclude that the limit of sin(1/x) as x approaches zero is also 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.