pls solve ! Lim x->0 sec(x)-1/ x sec(x) **i know the final answer but i dont know how to obtain it which is : lim x->0 1−cos(x)/x=0?

(sec(x)-1)/x multiply top and bottom by cos(x)

=(1-cos(x))/x since sec(x)=1/cos(x).

If you know that cos(x)= 1 -x^2/2!+x^4/4!-x^6/6!+....

then 1-cosx=x^2/2! -x^4/4! +x^6/6!-....divide by x

(1-cos(x))/x= x/2! - x^3/4!+.... and so

lim x->0[(1-cos(x))/x = 0

You could also use l'Hopital's Rule.

write all the sec(x)'s as 1/cos(x). then multiply the entire numerator and denominator by cos(x), and you'll get lim x->0 (1 - cos(x))/x. to solve this, apply l'hopital's rule and differentiate numerator and denominator. you get lim x->0 sin(x)/1 = 0 by subbing in, and we're done.

I want to direct your attention to my answer to your earlier limits question, where I gave two answers for a > 0 and a < 0. I should've just said the limit is -1/|a|, since we're talking about x-->0.