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
6 years, 4 months ago

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2 votes

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I assume cc is a finite value

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

Substitution of x=1x=1 must give an indeterminate form of 00\frac {0}{0} because denominator yields 00

So the numerator is equals to 00 when x=1a+b+1=0x=1 \Rightarrow a+b+1=0

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

4ax3+3bx2π(x1)cos(πx)+sin(πx) \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 00\frac {0}{0} as well because denominator yields 00 at x=1x=1, thus numerator must yield 00 as well

So, 4a+3b=04a+3b=0, solve the two equations gives a=3a = 3, b=4b=-4

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

36x224xπ[(x1)πsin(πx)+cos(πx)]+πcos(πx) \frac {36 x^{2} - 24 x } { \pi [ -(x-1) \pi \sin (\pi x) + \cos (\pi x) ] + \pi \cos (\pi x) }

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

Hence, a=3,b=4,c=6πa = 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 - 6 years, 4 months ago

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

A Former Brilliant Member - 6 years, 4 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 - 6 years, 4 months ago

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from where did u get this problem?

Vamsi Krishna Appili - 6 years, 4 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 - 6 years, 4 months ago

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How about We asume that limit when x1x \rightarrow 1 is 0/0. Then, calculate it using L'Hopital Theorem.. Sorry, I haven't tried that before..

Andrias Meisyal - 6 years, 4 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....:)

A Former Brilliant Member - 6 years, 4 months ago

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thank you very much for explanation :)

Andrias Meisyal - 6 years, 4 months ago

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@Andrias Meisyal no problem....

A Former Brilliant Member - 6 years, 4 months ago

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