How do you find the limit of #(x-pi/2)/cosx# as #x->pi/2#?

Answer 1

Use #cosx = -sin(x-pi/2)# and #lim_(thetararr0)theta/sin theta = 1#

#lim_(xrarrpi/2)(x-pi/2)/cosx = lim_(xrarrpi/2)(x-pi/2)/(-sin(x-pi/2)) #
With #theta = x-pi/2#, this is equivalent to
#lim_(theta rarr0) theta /(-sin theta) = - lim_(theta rarr0) theta /sin theta = -1#.
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Answer 2

To find the limit of (x - π/2)/cos(x) as x approaches π/2, we can use L'Hôpital's rule. Taking the derivative of the numerator and denominator, we get 1 and -sin(x) respectively. Substituting π/2 into the derivatives, we have 1 and -1. Therefore, the limit is 1/-1, which simplifies to -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.

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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