How do you determine the limit of #(pi/2)-(x)/(cos(x))# as x approaches pi/2?

Question

Charles Arocha · Accepted Answer

The limit does not exist, but

As #xrarrpi/2#, #cosxrarr0#, so the expression #x/cosx# either increases or decreases without bound depending on the direction of approach. However If the intended question is for the limit of#(pi/2-x)/cosx#, then use the trigonometric identity #cosx = sin(pi/2-x)# to get #lim_(xrarrpi/2) (pi/2-x)/cosx = lim_(xrarrpi/2) (pi/2-x)/sin(pi/2-x)#. With #u = pi/2-x# we have #lim_(urarr0)u/sinu = 1# by a fundamental trigonometric limit. That is, #lim_(xrarrpi/2) (pi/2-x)/cosx = 1#.

Elijah Abram · Answer

undefined To determine the limit of (pi/2) - (x)/(cos(x)) as x approaches pi/2, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately and then evaluate the limit of the resulting expression.

Differentiating the numerator, we get -1, and differentiating the denominator, we get -sin(x).

Now, substituting x = pi/2 into the differentiated expressions, we have -1 for the numerator and -sin(pi/2) = -1 for the denominator.

Therefore, the limit of (pi/2) - (x)/(cos(x)) as x approaches pi/2 is -1/-1, which simplifies to 1.