How do you use the Squeeze Theorem to find #lim sqrtx[1+ sin^2 (2π /x)]# as x approaches zero?

Answer 1

See explanation, below.

For all #x != 0#,
#-1 <= sin((2pi)/x) <= 1#, so
#0 <= sin^2((2pi)/x) <= 1#, and
#1 <= 1+sin^2((2pi)/x) <= 2#.
For all #x > 0#, we have #sqrtx > 0#, so we can multiply the inequality by #sqrtx# without changing the inequalities.
#sqrtx <= sqrtx (1+sin^2((2pi)/x)) <= 2sqrtx #.
Observe that #lim_(xrarr0^+)sqrtx = 0# and #lim_(xrarr0^+)2sqrtx = 0#.
Therefore, #lim_(xrarr0^+)sqrtx (1+sin^2((2pi)/x)) =0#
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Answer 2

To use the Squeeze Theorem to find the limit of the given expression as x approaches zero, we need to find two functions that "squeeze" the expression and have the same limit as x approaches zero.

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

Next, we can consider the function f(x) = x and g(x) = -x. It follows that f(x) ≤ sqrt(x[1+ sin^2 (2π /x)]) ≤ g(x) for all x ≠ 0.

Taking the limit as x approaches zero for all three functions, we have:

lim x → 0 (f(x)) = 0 lim x → 0 (g(x)) = 0

Therefore, by the Squeeze Theorem, the limit of sqrt(x[1+ sin^2 (2π /x)]) as x approaches zero is also 0.

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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.

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.

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.

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.

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