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I'm assuming you use L'hopital?
First check that it is an indeterminate form.
Such as \(0/0\) or \(\infty/\infty\)
Yea, it gets me \(\large \frac{1-1+0}{1-1}\)
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Which it indeed is...when you plug in x = 1 for the equation you get \(\large \frac{0}{0}\) So yes you can use L'Hopital
Second time around I got \[\LARGE \frac{-1+(\frac{1}{x})}{-5\pi sin(5 \pi x)}\]
With the \(\lim_{x \rightarrow 1}\) ofc
And that would mean another round of L'hopital?
Alright...and is.. Yes indeed^ lol another round we go!
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Alright, just wanted to make sure, thanks!
And it would turn out as \(\large \frac{1}{25\pi}\)?
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