How to calculate $\lim\limits_{x\to 2} \frac{1}{\sin{\pi x}}$ ?
For $x\to 2$, I get: $\frac{1}{0}$.
How to calculate $\lim\limits_{x\to 2} \frac{1}{\sin{\pi x}}$
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limits-without-lhopital
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0Are you sure that there is exist the limit ? – 2017-01-09
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3The limit does not exist. – 2017-01-09
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1the limit is undefined – 2017-01-09
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0What can you conclude if you get $1/0$ ? – 2017-01-09
2 Answers
2
We have
$\lim\limits_{x\to 2-0} \frac{1}{\sin{\pi x}}= - \infty$
and
$\lim\limits_{x\to 2+0} \frac{1}{\sin{\pi x}}= + \infty$.
0
$\displaystyle \lim_{x\to2}\frac{1}{\sin(\pi x)} = \lim_{t\to0}\frac{\pi t}{\pi t \sin(\pi t)}=\lim_{t\to0}\frac{1}{\pi t} = \lim_{x\to0}\frac{1}{t}$, which doesn't exist, because I could get $\epsilon = \frac{1}{\delta + 1}$ and for all $\delta$: $\frac{1}{\delta} > \frac{1}{\delta + 1}$.