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A limit problem

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..:)

Note by Leonardo Chandra
4 years, 6 months ago

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I 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, 6 months ago

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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. :)

Leonardo Chandra - 4 years, 6 months ago

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

Riya Gupta - 4 years, 6 months ago

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from where did u get this problem?

Vamsi Krishna Appili - 4 years, 6 months ago

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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..:)

Leonardo Chandra - 4 years, 6 months ago

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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..

Andrias Yuwantoko - 4 years, 6 months ago

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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....:)

Riya Gupta - 4 years, 6 months ago

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thank you very much for explanation :)

Andrias Yuwantoko - 4 years, 6 months ago

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@Andrias Yuwantoko no problem....

Riya Gupta - 4 years, 6 months ago

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