Limits in indeterminate form

lim (x-sinx)/(x-tanx)
x->0

I know that this is a [0/0] indeterminate form, and that the answer is (-1/2).

But how? I keep getting the answer as 0.( I use the L'Hospital rule and take the derivative of the numerator and denominator, and when I plug in 0 I get 0 as the answer.

You need to use L'Hopitals Rule again to get -1/2

After using L'Hopitals Rule the first time you get

(1-cosx)/(1-sec²x), which approaches the indeterminate form 0/0

Applying L'Hopitals Rule again yields

sinx/(-2sec²xtanx) which simplifies to -cosx/2sec²x, which approaches -1/2

Use L'Hopital's rule.
Derivative of numerator is 1-cos(x)
Derivative of denominator is 1-sec^2(x)
So given limit is same as lim of (1-cos(x))/(1-sec^2(x)) as x-->0
But this is again of the 0/0 form
So apply L'Hopital's rule again
Derivative of 1-cos(x) is sin(x)
Derivative of 1-sec^2(x)= -2sec(x)sec(x)tan(x)
So required limit is limit of sin(x)/[-2sec^2(x)tan(x)] as x-->0
[-cos^2(x)]/[2*cos(x)] by using 1/sec^2(x)=cos^2(x) and
sin(x)/tan(x)=1/cos(x)
which simplifies to limit of [-cos(x)]/2 as x-->0
and this limit is -1/2

The best way is to use l`Hospital rule...

lim(x-->0) (x-sinx) /(x -tanx) = lim(x-->0) (1-cosx)/(1-(secx^2) here we have the indeterminate form yet... than we can use the L`Hospital rule again:

= lim(x-->0) (0+sinx)/(0 -2secx*secxtanx) here we have the indeterminate form yet... repeat the L`Hospital rule again:

= lim(x-->0) (cosx)/( -2secxtanxsecxtanx -2secxsecxtanxtanx -2secxsecx(secx)^2) = 1/-2 = (-1/2) OK!

Note to find the derivative of -2secxsecxtancx we can use the rule product of three functions, that is:

(fgh)´ = f´gh + fg´h + fgh´ ...

When you use L'Hospital you get:

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

1-sec^2(x) = -tan^2(x) so

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

Still [0/0] so

lim(-sin(x)/2tan(x)sec^2(x))
x->0

which simplifies to

lim(-cos^3(x)/2)
x->0

which equals -1/2