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

$$Valueof\underset{x\to 0}{\lim }\frac{\cosh x-\cos x}{x\sin x}=?$$

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

$$byL-Hrule,i^{\prime }mgetting0.$$$$kindlycheck...$$

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

$$\underset{x\to 0}{\lim }\frac{\sinh x+\sin x}{\sin x+x\cos x}=\frac{\underset{x\to 0}{\lim }\frac{\sinh x}{\sin x}+1}{1+\underset{x\to 0}{\lim }\frac{x}{\sin x}\times \cos x}=\frac{1+1}{1+1}=1$$

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

$$thankssir.$$

Answered by tanmay last updated on 16/Apr/19

$$\underset{x\to 0}{\lim }\frac{\frac{e^x+e^{-x}}{2}-cosx}{xsinx}$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{e^x+e^{-x}-2cosx}{xsinx}$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...)+(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+..)-2(1-\frac{x^2}{2!}+\frac{x^4}{4!}-..)}{x(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)}$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{(2+\frac{2x^2}{2!}+\frac{2x^4}{4!}+...)-(2-\frac{2x^2}{2!}+\frac{2x^4}{4!}+...)}{x^2(1-\frac{x^2}{3!}+\frac{x^4}{5!}+..)}$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{2x^2+\frac{2x^6}{6!}+...}{x^2(1-\frac{x^2}{3!}+\frac{x^4}{5!})}$$$$=\frac{1}{2}\times \frac{2+0}{(1-0)}=1$$$$ormethod$$$$or\frac{1}{2}\underset{x\to 0}{\lim }\frac{e^x+e^{-x}-2cosx}{xsinx}\left(\frac{0}{0}\right)$$$$=\frac{1}{2}\underset{x\to 0}{\lim }\frac{e^x-e^{-x}+2sinx}{xcosx+sinx}$$$$=\frac{1}{2}\underset{x\to 0}{\lim }\frac{(e^x-1)-(e^{-x}-1)+2sinx}{xcosx+sinx}$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{\frac{e^x-1}{x}+\frac{e^{-x}-1}{-x}+2\frac{sinx}{x}}{cosx+\frac{sinx}{x}}$$$$\frac{1}{2}\times \frac{1+1+2\times 1}{1+1}$$$$\frac{1}{2}\times \left(\frac{4}{2}\right)=1$$$$$$$$$$$$$$$$$$$$$$$$$$$$\frac{1}{2}\underset{x\to 0}{\lim }$$

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

$$ThankUsir.$$

Commented by tanmay last updated on 16/Apr/19

$$mostwelcome$$

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