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What is the limit of 2x divided by tanx plus sinx as it approaches zero from the right?
Cheyne48
This answer will assume you understand basic concepts of limits. This is what I am interpreting your problem as:
lim x->0+ [(2x)/(tan(x) + sin(x))]
It is easy to see that simple substitution of 0 in for x will yield an indeterminate form 0/0. So, L'Hopital's rule will be applied to solve this limit. This rule states that an indeterminate form in a limit can still be solved for by deriving the top and bottom of the divided function and resolving for the limit. The "top" of this expression is 2x, and the "bottom" is tan(x) + sin(x). Deriving both top and bottom yields a new expression:
2/(sec2(x)+cos(x))
Substitution of 0 into this expression yields a determinate form, because sec2(0)=1/cos2(0)=1/1=1 and cos(0)=1, so the new "bottom" is 1+1=2. The general limit of this new expression is equal to the general limit of the original expression, so:
lim x->0 [2/(sec2(x) + cos(x))] = 2/2 = 1 = lim x->0 [(2x)/(tan(x) + sin(x))]
Since this is a general limit, the limit as x approaches zero from the left and right are equal, so they are both 1.
The answer is 1.
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