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Question Number 155353 by zakirullah last updated on 29/Sep/21
show that  lim_(x→0) ((sinx)/x) = 1
$$\boldsymbol{{show}}\:\boldsymbol{{that}}\:\:\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{0}} \frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\:=\:\mathrm{1} \\ $$
Answered by puissant last updated on 29/Sep/21
lim_(x→0) ((sinx)/x)=lim_(x→0) ((sinx−sin0)/(x−0))=f′(0)  f(x)=sinx→f′(x)=cosx  ⇒ lim_(x→0) ((sinx)/x)=f′(0)=cos(0)=1..
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinx}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinx}−{sin}\mathrm{0}}{{x}−\mathrm{0}}={f}'\left(\mathrm{0}\right) \\ $$$${f}\left({x}\right)={sinx}\rightarrow{f}'\left({x}\right)={cosx} \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinx}}{{x}}={f}'\left(\mathrm{0}\right)={cos}\left(\mathrm{0}\right)=\mathrm{1}.. \\ $$
Commented by zakirullah last updated on 29/Sep/21
sir nice but very difficult to understand ?  because some steps are skiped.  please do this very long process (simplification)
$$\mathrm{sir}\:\mathrm{nice}\:\mathrm{but}\:\mathrm{very}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{understand}\:? \\ $$$$\mathrm{because}\:\mathrm{some}\:\mathrm{steps}\:\mathrm{are}\:\mathrm{skiped}. \\ $$$$\mathrm{please}\:\mathrm{do}\:\mathrm{this}\:\mathrm{very}\:\mathrm{long}\:\mathrm{process}\:\left(\mathrm{simplification}\right) \\ $$
Commented by TheHoneyCat last updated on 29/Sep/21
This answer might not be perfectly written, but I can assure you it contains all the steps, as long as you know that dsin(x)/dx is cos... But I'm pretty sure it is not what you asked for. So do study it if you need, but know that there are no "less difficult" solutions.
Commented by zakirullah last updated on 30/Sep/21
ok sir i am understand.  thanks alot.  but lim_(x→0) ((sinx)/x) = ((sinx−sin0)/(x−0)) = f′(0)  oR lim_(x→0) ((sinx.sin0)/(x.o)) = f′(0)  which one is the most correct.
$$\mathrm{ok}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{am}\:\mathrm{understand}. \\ $$$$\mathrm{thanks}\:\mathrm{alot}. \\ $$$$\mathrm{but}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sinx}}{\mathrm{x}}\:=\:\frac{\mathrm{sinx}−\mathrm{sin0}}{\mathrm{x}−\mathrm{0}}\:=\:\mathrm{f}'\left(\mathrm{0}\right) \\ $$$$\mathrm{oR}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sinx}.\mathrm{sin0}}{\mathrm{x}.\mathrm{o}}\:=\:\mathrm{f}'\left(\mathrm{0}\right) \\ $$$$\mathrm{which}\:\mathrm{one}\:\mathrm{is}\:\mathrm{the}\:\mathrm{most}\:\mathrm{correct}. \\ $$
Answered by physicstutes last updated on 02/Oct/21
lim_(x→0) ((sin x)/x) = lim_(x→0) ((cos x)/1) = cos(0)=1
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:{x}}{{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}}{\mathrm{1}}\:=\:\mathrm{cos}\left(\mathrm{0}\right)=\mathrm{1} \\ $$

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