Question

lim x tends to 0 secx-1/tan^2x​

Answer 1

Answer:

Step-by-step explanation:

lim,x=>0 (secx -1)/tan^2 x; since tan^2x = sec^2x -1 =(secx +1)*(secx -1)

Remember, a^2 -b^2 = (a+b)*(a-b)

lim,x=>0 (secx -1)/((secx +1)*(secx -1)) = lim,x=>0(1/(secx +1))=1/(ses(0)+1)=1/(1+1)=1/2 , since sec(0)=1