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

Question

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

Charles Arocha · Accepted Answer

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#

Emma Aldridge · Answer

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#

Scarlett Allison · Answer

undefined To find the limit of cosx/(pi/2-x) as x approaches pi/2, we can use L'Hôpital's rule. Taking the derivative of both the numerator and denominator, we get -sinx/(-1), which simplifies to sinx. Evaluating this expression at x = pi/2, we find that the limit is equal to 1.

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

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