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

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Answer 1 · 2017-02-13T12:23:09

Because the expression evaluated at the limit results in an indeterminate form (specifically #0/0#), one should use L'Hôpital's rule .

Given: #lim_(xto"pi/2)cos(x)/(pi/2-x)=?#

Using L'Hôpital's rule , we compute the derivative of the numerator and the denominator:

#(d(cos(x)))/dx = -sin(x)#

#(d(pi/2-x))/dx = -1#

Assemble the new expression and evaluate at the limit:

#lim_(xto"pi/2)(-sin(x))/(-1)=1#

According to the rule, the original limit goes to the same value:

#lim_(xto"pi/2)cos(x)/(pi/2-x)=1#

Answer 2 · 2017-02-13T13:56:51

If you have not yet learned l'Hospital's Rule, then see below.

Use #lim_(thetararr0) sintheta/theta = 1# after using

#cosx = sin(pi/2-x)# (co-functiono identity from trigonometry).

#lim_(xrarrpi/s)cosx/(pi/2-x) = lim_(xrarrpi/2)sin(pi/2-x)/(pi/2-x)#

Note that: #lim(xrarrpi/2)(pi/2 -x) = 0#, so with #theta = pi/2-x#, we have

# = lim_(thetararr0)sintheta/theta = 1#