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Question Number 184891 by mathlove last updated on 13/Jan/23
lim_(x→π)  ((x−π)/(sinx))=?
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{{x}−\pi}{{sinx}}=? \\ $$
Answered by Mathspace last updated on 13/Jan/23
chang. x−π=t give  lim_(x→π) ((x−π)/(sinx))=lim_(t→0) (t/(sin(π+t)))  =−lim_(t→0) (t/(sint))=−1
$${chang}.\:{x}−\pi={t}\:{give} \\ $$$${lim}_{{x}\rightarrow\pi} \frac{{x}−\pi}{{sinx}}={lim}_{{t}\rightarrow\mathrm{0}} \frac{{t}}{{sin}\left(\pi+{t}\right)} \\ $$$$=−{lim}_{{t}\rightarrow\mathrm{0}} \frac{{t}}{{sint}}=−\mathrm{1} \\ $$
Answered by alephzero last updated on 13/Jan/23
lim_(x→π) ((x−π)/(sin x)) = lim_(x→π) (1/(cos x)) = (1/(−1)) = −1
$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{{x}−\pi}{\mathrm{sin}\:{x}}\:=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{1}}{−\mathrm{1}}\:=\:−\mathrm{1} \\ $$
Answered by TUN last updated on 14/Jan/23
=lim_(x→π)  ((−(π−x))/(sin(π−x)))  =−lim_(x→π)  ((π−x)/(sin(π−x)))  =−1
$$=\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{−\left(\pi−{x}\right)}{{sin}\left(\pi−{x}\right)} \\ $$$$=−\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\pi−{x}}{{sin}\left(\pi−{x}\right)} \\ $$$$=−\mathrm{1} \\ $$

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