Lim x-0 (tan x) / (x +(sin x))?

Question

Could you please show me step by step how to do this? I need to have work to show, and I looked it up on wolfram alpha, and it says the limit is 1/2, but their step-by-step uses L'Hospital's rule, and we haven't learned that yet, so I can't use that.

δοτζο · Accepted Answer

lim[x->0] sin(x) / cos(x)(x + sin(x))
lim[x->0] sin(x) / (x + sin(x)) [limits distribute over basically everything and cos(0) = 1]

If we assume that this limit is non-zero (we know it is, but pretending we don't), then we'll have that this is equal to
1 / lim[x->0] (x + sin(x)) / sin(x)

Then 
lim[x->0] (x + sin(x)) / sin(x)
lim[x->0] ((x / sin(x)) + 1)
1 + 1 = 2

So we have that the limit is 1/2

frank · Answer

lim   (tan x)/(x +(sin x))  (By l'hÃ´pital's rule) 
x-> 0

= lim (tan x)'/(x +sin x))'
x->0

= lim (sec x)/(1 + cos x)
x-> 0

= lim sec (0)/( 1 + cos 0)
x-> 0

=  1/ (1 + 1)

= 1/2 (answer)

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Lim x->0 (tan x) / (x +(sin x))?

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