How do you find the limit of #cosx/(1-sinx)# as x approaches pi/2+?

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Answer 1 · 2016-10-19T17:24:44

Use L'Hôpital's rule to discover that it approaches infinity as x approaches #pi/2#

If you try to evaluate the limit at #pi/2# you obtain the indeterminate form #0/0#; this means that L'Hôpital's rule applies.

To implement the rule, take the derivative of the numerator:

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

take the derivative of the denominator.

#(d{1 - sin(x)})/dx = -cos(x)#

Assemble this into a fraction:

#lim_(x->pi/2) (-sin(x))/(-cos(x))#

Please observe that the above is the tangent function:

#lim_(x->pi/2) tan(x)#

It is well known that the tangent function approaches infinity as x approaches #pi/2#, therefore, the original expression does the same thing.

Answer 2 · 2016-10-19T18:14:10

Multiply by #(1+sinx)/(1+sinx)#

#cosx/((1-sinx)) * ((1+sinx))/((1+sinx)) = (1+sinx)/cosx#

Now as #xrarr(pi/2)^+#, we have

#1+sinx rarr 2# and

#cosx rarr 0^-#

So

#lim_(xrarr (pi/2)^+) cosx/(1-sinx) = lim_(xrarr(pi/2)^+) (1+sinx)/cosx = -oo#