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

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Answer 1 · 2018-05-05T00:01:59

The answer
#lim_(xrarrpi)(x-pi)/(sinx)=lim_(xrarrpi)1/cosx=1/-1=-1#

show the steps

#lim_(xrarrpi)(x-pi)/(sinx)# Direct compensation product equal #(0/0)#

we must use L'Hopital's Rule

#lim_(xrarra)[f'(x)]/[g'(x)]# if the direct compensation product equal #(0/0)#

in your question
#f(x)=x-pi#

#f'(x)=1#

#g(x)=sinx#

#g'(x)=cosx#

#lim_(xrarrpi)(x-pi)/(sinx)=lim_(xrarrpi)1/cosx=1/-1=-1#