How do you use the Squeeze Theorem to find #lim(x-1)sin(pi/x-1) # as x approaches one?

Answer 1

#lim_(x rarr 1) (x-1)sin(pi/x - 1) = 0#

Firstly, we don't need the squeeze theorem because the function is continuous at values close to #x=1#, so the limit is #f(1)#, that being said, to use the squeeze theorem we must remember that
#-1 <= sin(theta) <= 1#
If we have #theta = pi/x - 1#, we have
#-1 <= sin(pi/x - 1) <= 1#
Multiplying both sides by #x-1# we have
#1 -x <= (x-1)sin(pi/x - 1) <= x - 1# for #x >0#

Since

#lim_(x rarr 1) 1 - x = lim_(x rarr 1) x - 1 = 0#

As per the squeeze theorem,

#lim_(x rarr 1) (x-1)sin(pi/x - 1) = 0#
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Answer 2

#lim_(xrarr1)(x-1)sin(pi/(x-1)) = 0#

#-1 <= sin(pi/(x-1)) <= 1# for all #x != 1#

The right-hand limit is 0.

For #x > 1#, we have #x-1 > 0# so we can multiply the inequality without changing the inequalities:
# -(x-1) <= (x-1)sin(pi/(x-1)) <= x-1# for all #x != 1#
#lim_(xrarr1^+) -(x-1) = 0# and #lim_(xrarr1^+) (x-1) = 0#
So #lim_(xrarr1^+) (x-1)sin(pi/(x-1)) = 0#

The left limit is 0.

For #x < 1#, we have #x-1 < 0# so when we multiply the inequality we must change the inequalities:
# -(x-1) >= (x-1)sin(pi/(x-1)) >= x-1# for all #x != 1#
#lim_(xrarr1^-) -(x-1) = 0# and #lim_(xrarr1^-) (x-1) = 0#
So #lim_(xrarr1^-) (x-1)sin(pi/(x-1)) = 0#

Consequently,

#lim_(xrarr1) (x-1)sin(pi/(x-1)) = 0#
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Answer 3

To use the Squeeze Theorem to find the limit of the function lim(x-1)sin(pi/x-1) as x approaches one, we can start by observing that the sine function is bounded between -1 and 1 for all values of x.

Next, we can rewrite the given function as sin(pi/(x-1)) and notice that as x approaches one, the denominator (x-1) approaches zero.

Using the Squeeze Theorem, we can establish two additional functions that bound the given function. Let's consider the functions f(x) = -1 and g(x) = 1.

Since sin(pi/(x-1)) is bounded between -1 and 1, we have -1 ≤ sin(pi/(x-1)) ≤ 1 for all x ≠ 1.

Therefore, by the Squeeze Theorem, as x approaches one, the limit of sin(pi/(x-1)) is also between -1 and 1.

Hence, the limit of the function lim(x-1)sin(pi/x-1) as x approaches one is 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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