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..:)

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

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.

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. :)

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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.

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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. :)

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yeah agreed......with the answer...:)

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from where did u get this problem?

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

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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..

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since denominator at x=1 tends to 0 then for the limit to exist ,numerator has to be 0 or else it will form indeterminate form....

0/0 is a basic form in limits....:)

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thank you very much for explanation :)

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