# Calculate limit x -x^3/6-sin x with range of sine .

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The problem does not specify if x approaches to a finite or infinite value. Since the problem indicates to solve the limit using the range of sine function, thus one can suppose that `x -> oo` .

Since the range of sine function is the interval `[-1,1]` yields:

`lim_(x->oo) (x - x^3/6 - sin x) > lim_(x->oo) (x - x^3/6 - 1)`

`lim_(x->oo) (x - x^3/6 - 1) = lim_(x->oo) (6x - x^3 - 6)/6 = oo`

Since` lim_(x->oo) (x - x^3/6 - sin x) > lim_(x->oo) (x - x^3/6 - 1)` yields `lim_(x->oo) (x - x^3/6 - sin x) = oo` .

**Hence, evaluating the llimit of function, considering `x->oo` , yields **`lim_(x->oo) (x - x^3/6 - sin x) = oo.`