Value-of-lim-x-0-cosh-x-cos-x-xsin-x-

Question Number 58016 by rahul 19 last updated on 16/Apr/19

V a l u e o f \lim_{x \to 0} \frac{\cosh x - \cos x}{x \sin x} = ?

Commented by rahul 19 last updated on 16/Apr/19

b y L - H r u l e, i^{'} m g e t t i n g 0 .

k i n d l y c h e c k \dots

Answered by mr W last updated on 16/Apr/19

\lim_{x \to 0} \frac{\sinh x + \sin x}{\sin x + x \cos x} = \frac{\lim_{x \to 0} \frac{\sinh x}{\sin x} + 1}{1 + \lim_{x \to 0} \frac{x}{\sin x} \times \cos x} = \frac{1 + 1}{1 + 1} = 1

Commented by rahul 19 last updated on 16/Apr/19

t h a n k s s i r .

Answered by tanmay last updated on 16/Apr/19

\lim_{x \to 0} \frac{\frac{e^{x} + e^{- x}}{2} - c o s x}{x s i n x}

\frac{1}{2} \lim_{x \to 0} \frac{e^{x} + e^{- x} - 2 c o s x}{x s i n x}

\frac{1}{2} \lim_{x \to 0} \frac{(1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots) + (1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + . .) - 2 (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - . .)}{x (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots)}

\frac{1}{2} \lim_{x \to 0} \frac{(2 + \frac{2 x^{2}}{2!} + \frac{2 x^{4}}{4!} + \dots) - (2 - \frac{2 x^{2}}{2!} + \frac{2 x^{4}}{4!} + \dots)}{x^{2} (1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} + . .)}

\frac{1}{2} \lim_{x \to 0} \frac{2 x^{2} + \frac{2 x^{6}}{6!} + \dots}{x^{2} (1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!})}

= \frac{1}{2} \times \frac{2 + 0}{(1 - 0)} = 1

o r m e t h o d

o r \frac{1}{2} \lim_{x \to 0} \frac{e^{x} + e^{- x} - 2 c o s x}{x s i n x} (\frac{0}{0})

= \frac{1}{2} \lim_{x \to 0} \frac{e^{x} - e^{- x} + 2 s i n x}{x c o s x + s i n x}

= \frac{1}{2} \lim_{x \to 0} \frac{(e^{x} - 1) - (e^{- x} - 1) + 2 s i n x}{x c o s x + s i n x}

\frac{1}{2} \lim_{x \to 0} \frac{\frac{e^{x} - 1}{x} + \frac{e^{- x} - 1}{- x} + 2 \frac{s i n x}{x}}{c o s x + \frac{s i n x}{x}}

\frac{1}{2} \times \frac{1 + 1 + 2 \times 1}{1 + 1}

\frac{1}{2} \times (\frac{4}{2}) = 1

\frac{1}{2} \lim_{x \to 0}

Commented by rahul 19 last updated on 16/Apr/19

T h a n k U s i r .

Commented by tanmay last updated on 16/Apr/19

m o s t w e l c o m e

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