The limit of [sin (SiNx)] / x, X → 0

0 / 0 type, using lobita rule
The derivation of the numerator and denominator gives cos (SiNx) * cosx / 1
Substitute x = 0 to get 1

Does the limit of 1 / sin (2-x) exist when x tends to 2? Why?

The limit is infinite
Note: positive infinity and negative infinity are infinity
The left and right limits are positive infinity, negative infinity and infinity

The limit definition of sin (x) △ x = 0 proves that x tends to be positive infinity

When x approaches positive infinity, sin (x) is bounded and √ x approaches infinity
sin（x）÷√x＝0

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