Limx-0 sinx/tanxsecx+tanx.sinx

Answer 1

Limit of sin(x)/(tan(x)*sec(x)+tan(x)*sin(x)) as x approaches 0

simplify first the expression:
sin(x)/(tan(x)*sec(x)+tan(x)*sin(x))
Note that tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x). Thus,
sin(x) / (sin(x)/cos(x) * 1/cos(x) + sin(x)/cos(x) * sin(x))
sin(x) / (sin(x)/cos^2(x) + sin^2(x)/cos(x)
1 / (1/cos^2(x) + sin(x)/cos(x))
or 1 / (sec^2 (x) + tan (x))
Then we find the limit as x -> 0. Substituting x = 0,
= 1 / (1/cos^2(0) + sin(0)/cos(0))
= 1 / (1/1^2 + 0/1)
= 1 / (1 + 0)
= 1 / 1
= 1

hope this helps~ `u`