Question

How to solve lim x-->0 of Sin3x / xcosx ?

I know my identities, I'm having trouble getting my sinx/x and 1- cosx /x

Answer 1

separate it in 2 fractions: sin 3x/x and 1/cos x
now multiply and divide by 3 in first fraction, so that angle of sin = 3x = denominator.
\(\large 3.\frac{sin3x}{3x}.\frac{1}{cosx}\)
can u do it further ?

Answer 2

I can simplify the sin3x and take the limit of that one (so 1) and factor out the three. Also, I can guess the final answer is 3, but how do you bring cosx over to the numerator ? Is it me who's stupid? I mean I can only think of putting it to the exponent -1

Answer 3

even if u put it to exponent -1, u get (cos x)^-1
when u put x=0
u get (cos 0)^-1 =1^-1 =1
so yes, u can do that also.

Answer 4

ahhh I thought I had to use the identity for cosx too, but it gives one anyway... thank you. :)

Answer 5

yup,welcome :)