I have a problem in limit that made me curious until now. The problem is about the limit and trigonometry. I don't know how to simplify it.. pls help me to solve it. thanks..:)

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TopNewestI assume \(c\) is a finite value

The key here is L'hôpital's Rule (3 times)

Substitution of \(x=1\) must give an indeterminate form of \(\frac {0}{0} \) because denominator yields \(0\)

So the numerator is equals to \(0\) when \(x=1 \Rightarrow a+b+1=0\)

Apply L'hôpital's Rule and chain rule, the limit becomes

\( \frac {4 a x^{3} + 3 b x^{2}} { \pi (x - 1) \cos (\pi x) + \sin (\pi x) } \)

And it must yield an indeterminate form of \(\frac {0}{0} \) as well because denominator yields \(0\) at \(x=1\), thus numerator must yield \(0 \) as well

So, \(4a+3b=0\), solve the two equations gives \(a = 3\), \(b=-4\)

Now substitute those two values into the limit and apply L'hôpital's Rule Rule once again

\( \frac {36 x^{2} - 24 x } { \pi [ -(x-1) \pi \sin (\pi x) + \cos (\pi x) ] + \pi \cos (\pi x) } \)

Substitute \(x=1\) gives \(c = - \frac {6}{\pi} \)

Hence, \(a = 3, b=-4, c = - \frac {6}{\pi} \)

By the way, I'm new to LaTeX, anyone know how to increase size of the mathematical symbols used? The fractions displayed are too small in standard font size. – Pi Han Goh · 4 years ago

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– Leonardo Chandra · 4 years ago

thanks for the answer, Pi Han..I never consider the using of 0/0 to evaluate the value of numerator, then apply L'hopital more than twice to find the answer. Anyway, thanks for the great answer..I need to learn more again about the L'Hopital rule. :)Log in to reply

– Riya Gupta · 4 years ago

yeah agreed......with the answer...:)Log in to reply

from where did u get this problem? – Vamsi Krishna Appili · 4 years ago

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– Leonardo Chandra · 4 years ago

I got this problem from my calculus book. this problem is about limit problems..an introduction to improper form of limit..:)Log in to reply

How about We asume that limit when \(x \rightarrow 1 \) is 0/0. Then, calculate it using L'Hopital Theorem.. Sorry, I haven't tried that before.. – Andrias Yuwantoko · 4 years ago

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0/0 is a basic form in limits....:) – Riya Gupta · 4 years ago

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– Andrias Yuwantoko · 4 years ago

thank you very much for explanation :)Log in to reply

– Riya Gupta · 4 years ago

no problem....Log in to reply