Factorials! and a hyperbolic function

Calculus Level 4

lim x ( ( x + 2 ) ! cosh ( 1 x ) x 2 ( x ! + 1 ) ) x = ? \large \lim_{x \to \infty} \left(\frac {(x+2)!\cosh \left(\frac 1{\sqrt x} \right)}{x^2(x! + 1)} \right)^x = \ ?

Round your answer to 4 decimal places.

Notation : cosh (x) denotes the hyperbolic cosine .


The answer is 33.1155.

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1 solution

Andrea Virgillito
Feb 11, 2017

I have simplified by x ! x 2 x!x^2 : lim x ( ( 1 + 2 x ) ( 1 + 1 x ) cosh ( 1 x ) ( 1 + 1 x ! ) ) x \lim_{x \mapsto \infty} (\frac{(1+\frac{2}{x})(1+\frac{1}{x})\cosh (\frac{1}{\sqrt{x}})}{(1+\frac{1}{x!})})^x and then I have subdivided the limit in three parts: lim x ( 1 + 2 x ) ( 1 + 1 x ) x \lim_{x \mapsto \infty} (1+\frac{2}{x})(1+\frac{1}{x})^x , lim x ( 1 ( 1 + 1 x ! ) ) x \lim_{x \mapsto \infty} (\frac{1}{(1+\frac{1}{x!})})^x and lim x ( cosh ( 1 x ) ) x \lim_{x \mapsto \infty} (\cosh (\frac{1}{\sqrt{x}}))^x . To solve the first one you just have to expand the multiplication to obtain a normal neperian limit which results e 3 e^3 , in the indeterminate forme of the second limit predominates the 1 thus 1 1^ \infty = 1 finally the third one can be solved using the fact that cosh 2 ( x ) sinh 2 ( x ) = 1 \cosh^2(x)-\sinh^2(x)=1 and that lim x 0 s i n h ( x ) x = 1 \lim_{x \to 0} \frac{sinh(x)}{x}=1 that leads to lim x ( cosh ( 1 x ) ) x \lim_{x \mapsto \infty} (\cosh (\frac{1}{\sqrt{x}}))^x = e \sqrt{e} .

e 3 e^3 1 *1* e \sqrt{e} = e 7 2 e^\frac{7}{2} . The approximative value of e 7 2 e^\frac{7}{2} is 33.1155

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