How do you find the limit of $\frac{x - π}{sin x}$ $(x-pi)/(sinx)$ as x approaches pi?

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Answer 1 · 2018-05-05T00:01:59

The answer
$lim_(xrarrpi)(x-pi)/(sinx)=lim_(xrarrpi)1/cosx=1/-1=-1$

show the steps

$lim_(xrarrpi)(x-pi)/(sinx)$ Direct compensation product equal $(0/0)$

we must use L'Hopital's Rule

$lim_(xrarra)[f'(x)]/[g'(x)]$ if the direct compensation product equal $(0/0)$

in your question
$f(x)=x-pi$

$f'(x)=1$

$g(x)=sinx$

$g'(x)=cosx$

$lim_(xrarrpi)(x-pi)/(sinx)=lim_(xrarrpi)1/cosx=1/-1=-1$

How do you find the limit of (x-pi)/(sinx)x−πsinx (x-pi)/(sinx) as x approaches pi?