help? limit as x-->0 of x(tanx) /( 1-cosx)?

Question

so far ive gotten it to lim as x--> 0 xsinx / ((1-cosx)(cosx))

where do i go from here? thanks for your help!

Answer 1

This limit is of the indeterminant form 0/0 so L'Hospital's Rule applies. The limit of the quotient is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

Limit as x→0 of { x(tanx) /(1 - cosx) }

= Limit as x→0 { [tanx + x(sec²x)] / sinx }

This is still of indeterminant form. Repeat.

= Limit as x→0 { [sec²x + sec²x + x(2sec²x)(tanx)] / cosx }

= 2/1 = 2

Answer 2

Without L'Hopital's:

multiply top and bottom by 1+cos(x). Now you have:

lim x(tan x)(1+cos x) / (1-cos^2(x))
x->0

= x (tan x + sin x) / (sin^2 x)

= x / sin(x) * (tan x + sin x) / sin x

= x / sin(x) * (sec x + 1)

It is well known that

lim sin(n) / n = 1
n->0

So the limit is equal to 1 * (1+1) = 2.

Answer 3

Apply L'Hospitals rule.Too long to explain.Search on google or just consult wikipedia.