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I have simplified by x ! x 2 : lim x ↦ ∞ ( ( 1 + x ! 1 ) ( 1 + x 2 ) ( 1 + x 1 ) cosh ( x 1 ) ) x and then I have subdivided the limit in three parts: lim x ↦ ∞ ( 1 + x 2 ) ( 1 + x 1 ) x , lim x ↦ ∞ ( ( 1 + x ! 1 ) 1 ) x and lim x ↦ ∞ ( cosh ( x 1 ) ) x . To solve the first one you just have to expand the multiplication to obtain a normal neperian limit which results e 3 , in the indeterminate forme of the second limit predominates the 1 thus 1 ∞ = 1 finally the third one can be solved using the fact that cosh 2 ( x ) − sinh 2 ( x ) = 1 and that lim x → 0 x s i n h ( x ) = 1 that leads to lim x ↦ ∞ ( cosh ( x 1 ) ) x = e .
e 3 ∗ 1 ∗ e = e 2 7 . The approximative value of e 2 7 is 33.1155