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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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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## Comments

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

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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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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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