Evaluate, if limit doesn't exist explain: limx-->∞ sin(x-1 / 2+x^2)

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Supposing that you need to evaluate the limit `lim_(x-gtoo) sin((x-1)/(2+x^2)), ` hence you need to evaluate the sine of the limit of fraction `((x-1)/(2+x^2))`  such that:

`lim_(x-gtoo) sin((x-1)/(2+x^2)) = sin lim_(x-gtoo)((x-1)/(2+x^2))`

You need to force x factor to numerator and `x^2`  to denominator such that:

`sin lim_(x-gtoo)(x(1-1/x))/(x^2(2/x^2+1))`

`sin lim_(x-gtoo)((1-1/x))/(x(2/x^2+1)) = sin (1-0)/(oo(0+1))`

`sin lim_(x-gtoo)((1-1/x))/(x(2/x^2+1)) = sin (1/oo)`

lim_(x->oo) sin((x-1)/(2+x^2)) =

`sin lim_(x-gtoo)((1-1/x))/(x(2/x^2+1)) = sin 0`

`sin lim_(x-gtoo)((1-1/x))/(x(2/x^2+1)) = 0`

Hence, evaluating the limit of the function `sin((x-1)/(2+x^2))`  yields `lim_(x-gtoo) sin((x-1)/(2+x^2)) = 0` .

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