Calculate the limites: Lim x→π/2 (tan(x))/[3(x-π/2)] and Lim x→1 [(x^3-3x^2+6x-4)/(3x^3-x^2+9x+7)]π = pi



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Posted on (Answer #1)

With more than one question you need to make separate posts.

To find the limit `lim_{x->pi/2}{tan x}/{3(x-pi/2)}` , we can use L'Hopital's rule.

Start with the limit

`lim_{x->pi/2}{tan x}/{3(x-pi/2)}`   take out the factor 3

`=1/3lim_{x->pi/2}{tan x}/{x-pi/2}`    take derivative of numerator and denominator

`=1/3lim_{x->pi/2}{sec^2 x}/1`   replace sec with 1/cos

`=1/3 lim_{x->pi/2} 1/{cos^2x}`   but this is 1/0


The limit is undefined.


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