This is an indeterminate form and I think I should use the fact that $x-2/sin(x-2) = 1 ;$but idk how to vày that.

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$$lim_x o 2 fracx^2-4sin(x-2) = lim_x o 2 left( (x+2) fracx-2sin(x-2) ight) = left( lim_x o2 (x+2) ight) left( lim_x o2 fracx-2sin(x-2) ight)$$

The second limit may be written as $displaystylelim_u o0 frac u sin u$ và you"ve probably seen that limit evaluated by squeezing.



While $alphasim 0$, for example when $alpha o 0$, then $sin(alpha)sim alpha$. Here $x o 2$ so we have $$(x-2) o 0$$



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