# 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

### 1 Answer | Add Yours

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

=undefined.

**The limit is undefined.**

**Sources:**