How is #NH_3 # acting in the reaction #H^++:NH_3 ->[H:NH_3]^+#according to the Lewis definition?

Answer 1
#NH_3# is acting as an electron donor, so by definition it is a Lewis base.
The lone pair of elctrons from the #N#-atom is (mainly) donated to, are shared with, the proton (#H^+#), after which all four #H#'s fill up a nice tetrahedron around the #N#.
The representation of #[H:NH_3]^+# is not really correct, as all #H#-atoms are completely equal, and the electron deficit (the overall +) is shared around the whole #[NH_4]^+# ammonium -ion.
If the ammonium-ion is dissociated (e.g. in hot alkalic surroundings), there is no telling which of the four #H#'s will leave.
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Answer 2

NH₃ acts as a Lewis base in the reaction by donating a pair of electrons from its nitrogen atom to the proton (H⁺). This donation forms a coordinate covalent bond between the nitrogen atom of NH₃ and the proton, resulting in the formation of a positively charged ammonium ion, [H:NH₃]⁺.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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