Question

`lim_(x->0)tan(x)/(x+sin(x))`?

Answer 1

`= 1/2`

Explanation:

`lim_(x->0)tan(x)/(x+sin(x))`

Using Taylor Expansions:

`=lim_(x->0)(x + x^3/3 + mathcal O x^5)/(x+(x - x^3/(3!) + mathcal O x^5))`

`=lim_(x->0)(x + x^3/3 + mathcal O x^5)/(2x- x^3/(3!) + mathcal O x^5)`

`=lim_(x->0)(1 + x^2/3 + mathcal O x^4)/(2- x^2/(3!) + mathcal O x^4)`

`=1/2`

Answer 2

`tanx/(x+sinx) = (sinx/cosx)/(x+sinx)`

`= sinx/(cosx(x+sinx))`

`= (sinx/x)/(cosx(1+sinx/x))` for all `x != 0`

Now use `lim_(xrarr0) sinx/x =1` to get

`lim_(xrarr0)tanx/(x+sinx) = lim_(xrarr0) (sinx/x)/(cosx(1+sinx/x))`

`= 1/(x(1+1)) = 1/2`